What's the value of $sumlimits_{k=1}^{infty}frac{k^2}{k!}$?
$begingroup$
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $sumlimits_{k=1}^{infty}frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$sum_{k=1}^{infty}frac{k^2}{k!}?$$
calculus real-analysis sequences-and-series exponential-function factorial
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
$endgroup$
add a comment |
$begingroup$
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $sumlimits_{k=1}^{infty}frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$sum_{k=1}^{infty}frac{k^2}{k!}?$$
calculus real-analysis sequences-and-series exponential-function factorial
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
$endgroup$
add a comment |
$begingroup$
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $sumlimits_{k=1}^{infty}frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$sum_{k=1}^{infty}frac{k^2}{k!}?$$
calculus real-analysis sequences-and-series exponential-function factorial
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
$endgroup$
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $sumlimits_{k=1}^{infty}frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$sum_{k=1}^{infty}frac{k^2}{k!}?$$
calculus real-analysis sequences-and-series exponential-function factorial
calculus real-analysis sequences-and-series exponential-function factorial
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
edited Apr 16 '17 at 16:42
Martin Sleziak
44.7k10118272
44.7k10118272
asked Jun 8 '11 at 15:33
JackJack
1
asked Jun 8 '11 at 15:33
JackJack
1
asked Jun 8 '11 at 15:33
JackJack
1
1
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
$endgroup$
add a comment |
$begingroup$
The value of $T_n := displaystylesum_{k=1}^{infty} frac{k^n}{k!}$ is $B_n cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$begin{align}
T_{n+1} = sum_{k=1}^{infty} frac{k^{n+1}}{k!} &= sum_{k=0}^{infty} frac{(k+1)^n}{k!}
\
&= sum_{k=0}^{infty} frac{1}{k!} sum_{j=0}^n {n choose j} k^j
\
&= sum_{j=0}^n {n choose j} sum_{k=1}^{infty} frac{k^j}{k!}
\
&= sum_{j=0}^{n} {n choose j} T_j
end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
$endgroup$
add a comment |
$begingroup$
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=sum limits_{n=0}^infty frac{x^n}{n!}$, apply $frac{d}{dx}xfrac{d}{dx}$ to both sides, and evaluate at $x=1$.
edited Dec 22 '18 at 1:30
Shaun
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
$endgroup$
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
add a comment |
$begingroup$
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $xmapstoexp(x)$.
answered Jun 8 '11 at 15:38
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
A basic technique in real (complex) analysis is term by term differentiation of power series:
$$
e^z=sum_{k=0}^inftyfrac{z^k}{k!},quad e^z=(e^z)'=sum_{k=1}^infty kcdotfrac{z^{k-1}}{k!},quad
e^z=(e^z)''=sum_{k=1}^infty k(k-1)frac{z^{k-2}}{k!}.
$$
Evaluating at $z=1$, one immediately has
$$
e=sum_{k=0}^inftyfrac{1}{k!},quad e=sum_{k=1}^infty kcdotfrac{1}{k!},quad
e=sum_{k=1}^infty (k^2-k)frac{1}{k!}.
$$
Combining the second and third equalities, we have the answer.
answered Mar 15 '16 at 2:47
JackJack
1
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f44113%2fwhats-the-value-of-sum-limits-k-1-infty-frack2k%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
$endgroup$
add a comment |
$begingroup$
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
$endgroup$
add a comment |
$begingroup$
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
$endgroup$
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
answered Jun 8 '11 at 15:38
Luboš MotlLuboš Motl
7,05611625
7,05611625
add a comment |
add a comment |
$begingroup$
The value of $T_n := displaystylesum_{k=1}^{infty} frac{k^n}{k!}$ is $B_n cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$begin{align}
T_{n+1} = sum_{k=1}^{infty} frac{k^{n+1}}{k!} &= sum_{k=0}^{infty} frac{(k+1)^n}{k!}
\
&= sum_{k=0}^{infty} frac{1}{k!} sum_{j=0}^n {n choose j} k^j
\
&= sum_{j=0}^n {n choose j} sum_{k=1}^{infty} frac{k^j}{k!}
\
&= sum_{j=0}^{n} {n choose j} T_j
end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
$endgroup$
add a comment |
$begingroup$
The value of $T_n := displaystylesum_{k=1}^{infty} frac{k^n}{k!}$ is $B_n cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$begin{align}
T_{n+1} = sum_{k=1}^{infty} frac{k^{n+1}}{k!} &= sum_{k=0}^{infty} frac{(k+1)^n}{k!}
\
&= sum_{k=0}^{infty} frac{1}{k!} sum_{j=0}^n {n choose j} k^j
\
&= sum_{j=0}^n {n choose j} sum_{k=1}^{infty} frac{k^j}{k!}
\
&= sum_{j=0}^{n} {n choose j} T_j
end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
$endgroup$
add a comment |
$begingroup$
The value of $T_n := displaystylesum_{k=1}^{infty} frac{k^n}{k!}$ is $B_n cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$begin{align}
T_{n+1} = sum_{k=1}^{infty} frac{k^{n+1}}{k!} &= sum_{k=0}^{infty} frac{(k+1)^n}{k!}
\
&= sum_{k=0}^{infty} frac{1}{k!} sum_{j=0}^n {n choose j} k^j
\
&= sum_{j=0}^n {n choose j} sum_{k=1}^{infty} frac{k^j}{k!}
\
&= sum_{j=0}^{n} {n choose j} T_j
end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
$endgroup$
The value of $T_n := displaystylesum_{k=1}^{infty} frac{k^n}{k!}$ is $B_n cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$begin{align}
T_{n+1} = sum_{k=1}^{infty} frac{k^{n+1}}{k!} &= sum_{k=0}^{infty} frac{(k+1)^n}{k!}
\
&= sum_{k=0}^{infty} frac{1}{k!} sum_{j=0}^n {n choose j} k^j
\
&= sum_{j=0}^n {n choose j} sum_{k=1}^{infty} frac{k^j}{k!}
\
&= sum_{j=0}^{n} {n choose j} T_j
end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
edited Jul 18 '12 at 3:33
J. M. is not a mathematician
61.2k5151290
61.2k5151290
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
answered Jun 8 '11 at 16:26
JavaManJavaMan
11.1k12755
11.1k12755
add a comment |
add a comment |
$begingroup$
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=sum limits_{n=0}^infty frac{x^n}{n!}$, apply $frac{d}{dx}xfrac{d}{dx}$ to both sides, and evaluate at $x=1$.
edited Dec 22 '18 at 1:30
Shaun
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
$endgroup$
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
add a comment |
$begingroup$
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=sum limits_{n=0}^infty frac{x^n}{n!}$, apply $frac{d}{dx}xfrac{d}{dx}$ to both sides, and evaluate at $x=1$.
edited Dec 22 '18 at 1:30
Shaun
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
$endgroup$
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
add a comment |
$begingroup$
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=sum limits_{n=0}^infty frac{x^n}{n!}$, apply $frac{d}{dx}xfrac{d}{dx}$ to both sides, and evaluate at $x=1$.
edited Dec 22 '18 at 1:30
Shaun
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
$endgroup$
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=sum limits_{n=0}^infty frac{x^n}{n!}$, apply $frac{d}{dx}xfrac{d}{dx}$ to both sides, and evaluate at $x=1$.
edited Dec 22 '18 at 1:30
Shaun
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
edited Dec 22 '18 at 1:30
Shaun
9,241113684
edited Dec 22 '18 at 1:30
Shaun
9,241113684
edited Dec 22 '18 at 1:30
Shaun
9,241113684
9,241113684
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
answered Jun 8 '11 at 15:46
Ross MillikanRoss Millikan
297k23198371
297k23198371
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
add a comment |
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
2
2
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
$begingroup$
$frac{d}{dx}xfrac{d}{dx}$, typo?
$endgroup$
– Jack
Jun 8 '11 at 17:46
14
14
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
$begingroup$
It means: differentiate, then multiply by $x$, then differentiate.
$endgroup$
– GEdgar
Jun 8 '11 at 17:54
add a comment |
$begingroup$
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $xmapstoexp(x)$.
answered Jun 8 '11 at 15:38
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $xmapstoexp(x)$.
answered Jun 8 '11 at 15:38
DidDid
248k23224463
$endgroup$
add a comment |
$begingroup$
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $xmapstoexp(x)$.
answered Jun 8 '11 at 15:38
DidDid
248k23224463
$endgroup$
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $xmapstoexp(x)$.
answered Jun 8 '11 at 15:38
DidDid
248k23224463
answered Jun 8 '11 at 15:38
DidDid
248k23224463
answered Jun 8 '11 at 15:38
DidDid
248k23224463
answered Jun 8 '11 at 15:38
DidDid
248k23224463
248k23224463
add a comment |
add a comment |
$begingroup$
A basic technique in real (complex) analysis is term by term differentiation of power series:
$$
e^z=sum_{k=0}^inftyfrac{z^k}{k!},quad e^z=(e^z)'=sum_{k=1}^infty kcdotfrac{z^{k-1}}{k!},quad
e^z=(e^z)''=sum_{k=1}^infty k(k-1)frac{z^{k-2}}{k!}.
$$
Evaluating at $z=1$, one immediately has
$$
e=sum_{k=0}^inftyfrac{1}{k!},quad e=sum_{k=1}^infty kcdotfrac{1}{k!},quad
e=sum_{k=1}^infty (k^2-k)frac{1}{k!}.
$$
Combining the second and third equalities, we have the answer.
answered Mar 15 '16 at 2:47
JackJack
1
$endgroup$
add a comment |
$begingroup$
A basic technique in real (complex) analysis is term by term differentiation of power series:
$$
e^z=sum_{k=0}^inftyfrac{z^k}{k!},quad e^z=(e^z)'=sum_{k=1}^infty kcdotfrac{z^{k-1}}{k!},quad
e^z=(e^z)''=sum_{k=1}^infty k(k-1)frac{z^{k-2}}{k!}.
$$
Evaluating at $z=1$, one immediately has
$$
e=sum_{k=0}^inftyfrac{1}{k!},quad e=sum_{k=1}^infty kcdotfrac{1}{k!},quad
e=sum_{k=1}^infty (k^2-k)frac{1}{k!}.
$$
Combining the second and third equalities, we have the answer.
answered Mar 15 '16 at 2:47
JackJack
1
$endgroup$
add a comment |
$begingroup$
A basic technique in real (complex) analysis is term by term differentiation of power series:
$$
e^z=sum_{k=0}^inftyfrac{z^k}{k!},quad e^z=(e^z)'=sum_{k=1}^infty kcdotfrac{z^{k-1}}{k!},quad
e^z=(e^z)''=sum_{k=1}^infty k(k-1)frac{z^{k-2}}{k!}.
$$
Evaluating at $z=1$, one immediately has
$$
e=sum_{k=0}^inftyfrac{1}{k!},quad e=sum_{k=1}^infty kcdotfrac{1}{k!},quad
e=sum_{k=1}^infty (k^2-k)frac{1}{k!}.
$$
Combining the second and third equalities, we have the answer.
answered Mar 15 '16 at 2:47
JackJack
1
$endgroup$
A basic technique in real (complex) analysis is term by term differentiation of power series:
$$
e^z=sum_{k=0}^inftyfrac{z^k}{k!},quad e^z=(e^z)'=sum_{k=1}^infty kcdotfrac{z^{k-1}}{k!},quad
e^z=(e^z)''=sum_{k=1}^infty k(k-1)frac{z^{k-2}}{k!}.
$$
Evaluating at $z=1$, one immediately has
$$
e=sum_{k=0}^inftyfrac{1}{k!},quad e=sum_{k=1}^infty kcdotfrac{1}{k!},quad
e=sum_{k=1}^infty (k^2-k)frac{1}{k!}.
$$
Combining the second and third equalities, we have the answer.
answered Mar 15 '16 at 2:47
JackJack
1
answered Mar 15 '16 at 2:47
JackJack
1
answered Mar 15 '16 at 2:47
JackJack
1
answered Mar 15 '16 at 2:47
JackJack
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f44113%2fwhats-the-value-of-sum-limits-k-1-infty-frack2k%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
