If $sum a_k$ converges, does $sum(a_{k+1}- 2 a_{k+3})$ converge as well?
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If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?
My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$
Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.
real-analysis sequences-and-series proof-verification
$endgroup$
add a comment |
$begingroup$
If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?
My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$
Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.
real-analysis sequences-and-series proof-verification
$endgroup$
2
$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
1
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33
add a comment |
$begingroup$
If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?
My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$
Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.
real-analysis sequences-and-series proof-verification
$endgroup$
If $sum_{k=0}^infty a_k$ is convergent with value $s$, what about $sum_{k=0}^infty b_k$ where $b_k=a_{k+1}- 2 a_{k+3}$?
My reasoning:
$$sum_{k=0}^infty b_k =lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} sum_{k=0}^na_{k+1}-2a_{k+3}
$$
Within the sum we are only dealing with finitely many terms, we can split up the sum and take the limit afterwards:
$$ lim_{n rightarrow infty} sum_{k=0}^n b_k=lim_{n rightarrow infty} left( sum_{k=0}^n a_{k+1}- sum_{k=0}^n 2a_{k+3} right) $$
Now we want to get $s$ in here, the value of our sum, we need to do some index juggling, since our sum is not of the right form yet.
$$lim_{n rightarrow infty} left( sum_{k=0}^n a_{k} - a_0- 2sum_{k=0}^n a_{k} +2a_0 + 2a_1+2a_2right) $$
Here we applied an index shift, but then we need to compensate for the terms that we added to the sum, we finally apply the limit and get:
$$sum_{k=0}^infty b_k=s +a_0-2s+2a_1+2a_2= a_0+2a_1+2a_2 -s $$
Did that all make sense, is my reasoning correct?
conclusion: it converges as we computed the exact value.
real-analysis sequences-and-series proof-verification
real-analysis sequences-and-series proof-verification
edited Dec 3 '18 at 11:32
Did
247k23222458
247k23222458
asked Dec 2 '18 at 13:07
Wesley StrikWesley Strik
1,653423
1,653423
2
$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
1
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33
add a comment |
2
$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
1
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33
2
2
$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
1
1
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33
add a comment |
1 Answer
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I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.
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add a comment |
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I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.
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I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.
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add a comment |
$begingroup$
I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.
$endgroup$
I will write this in the answer section so that the question does not remain open.
The reasoning looks fine and correct to me.
answered Dec 4 '18 at 17:55
CosminCosmin
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$begingroup$
If $sum x_n = X in mathbb{R}$ and $sum y_n = Y in mathbb{R}$, then $sum (x_n + y_n) =X+ Y$, so yes, the reasoning is correct.
$endgroup$
– Cosmin
Dec 2 '18 at 13:40
1
$begingroup$
Looks fine to me
$endgroup$
– Shubham Johri
Dec 2 '18 at 14:22
$begingroup$
Please provide your answers in the answer section. This way the question does not stay open.
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:31
$begingroup$
math.meta.stackexchange.com/questions/1559/…
$endgroup$
– Wesley Strik
Dec 3 '18 at 11:33