Does $sum_{n=1}^{infty}frac{sinleft(frac{1}{sqrt n}right)}{2n-1}$ converge or diverge?
1
$begingroup$
I tried solving this but couldn't. I tried a sorta hacky approach and that was substituting $u = frac{1}{sqrt n}$ and made my way to a comparison which resulted in $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ but then I noticed that I didn't changed my summation bounds since it was from $n=1$ to $n to infty$ and I now think I should have changed it to be suitable for $u$ but I guessed it'd be wrong and bizzare so I didn't even attempt it. I know it's hacky and this wasn't really taught by my professors or anything, I was just giving it a go. What's the correct way to approach this?
sequences-and-series
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
$endgroup$
|
show 1 more comment
1
$begingroup$
I tried solving this but couldn't. I tried a sorta hacky approach and that was substituting $u = frac{1}{sqrt n}$ and made my way to a comparison which resulted in $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ but then I noticed that I didn't changed my summation bounds since it was from $n=1$ to $n to infty$ and I now think I should have changed it to be suitable for $u$ but I guessed it'd be wrong and bizzare so I didn't even attempt it. I know it's hacky and this wasn't really taught by my professors or anything, I was just giving it a go. What's the correct way to approach this?
sequences-and-series
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
$endgroup$
$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
$endgroup$
– Decaf-Math
Dec 5 '18 at 23:31
$begingroup$
I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:36
1
$begingroup$
@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
$endgroup$
– gimusi
Dec 5 '18 at 23:39
$begingroup$
@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
$endgroup$
– gimusi
Dec 5 '18 at 23:41
1
$begingroup$
@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
$endgroup$
– gimusi
Dec 5 '18 at 23:49
|
show 1 more comment
1
1
1
$begingroup$
I tried solving this but couldn't. I tried a sorta hacky approach and that was substituting $u = frac{1}{sqrt n}$ and made my way to a comparison which resulted in $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ but then I noticed that I didn't changed my summation bounds since it was from $n=1$ to $n to infty$ and I now think I should have changed it to be suitable for $u$ but I guessed it'd be wrong and bizzare so I didn't even attempt it. I know it's hacky and this wasn't really taught by my professors or anything, I was just giving it a go. What's the correct way to approach this?
sequences-and-series
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
$endgroup$
I tried solving this but couldn't. I tried a sorta hacky approach and that was substituting $u = frac{1}{sqrt n}$ and made my way to a comparison which resulted in $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ but then I noticed that I didn't changed my summation bounds since it was from $n=1$ to $n to infty$ and I now think I should have changed it to be suitable for $u$ but I guessed it'd be wrong and bizzare so I didn't even attempt it. I know it's hacky and this wasn't really taught by my professors or anything, I was just giving it a go. What's the correct way to approach this?
sequences-and-series
sequences-and-series
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
edited Dec 5 '18 at 23:44
gimusi
92.8k84494
92.8k84494
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
asked Dec 5 '18 at 23:26
Eyad H.Eyad H.
382111
382111
$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
$endgroup$
– Decaf-Math
Dec 5 '18 at 23:31
$begingroup$
I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:36
1
$begingroup$
@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
$endgroup$
– gimusi
Dec 5 '18 at 23:39
$begingroup$
@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
$endgroup$
– gimusi
Dec 5 '18 at 23:41
1
$begingroup$
@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
$endgroup$
– gimusi
Dec 5 '18 at 23:49
|
show 1 more comment
$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
$endgroup$
– Decaf-Math
Dec 5 '18 at 23:31
$begingroup$
I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:36
1
$begingroup$
@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
$endgroup$
– gimusi
Dec 5 '18 at 23:39
$begingroup$
@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
$endgroup$
– gimusi
Dec 5 '18 at 23:41
1
$begingroup$
@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
$endgroup$
– gimusi
Dec 5 '18 at 23:49
$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
$endgroup$
– Decaf-Math
Dec 5 '18 at 23:31
$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
$endgroup$
– Decaf-Math
Dec 5 '18 at 23:31
$begingroup$
I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:36
$begingroup$
I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:36
1
1
$begingroup$
@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
$endgroup$
– gimusi
Dec 5 '18 at 23:39
$begingroup$
@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
$endgroup$
– gimusi
Dec 5 '18 at 23:39
$begingroup$
@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
$endgroup$
– gimusi
Dec 5 '18 at 23:41
$begingroup$
@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
$endgroup$
– gimusi
Dec 5 '18 at 23:41
1
1
$begingroup$
@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
$endgroup$
– gimusi
Dec 5 '18 at 23:49
$begingroup$
@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
$endgroup$
– gimusi
Dec 5 '18 at 23:49
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
HINT
We have that for $n$ large
$$frac{sinleft(frac{1}{sqrt n }right)}{2n-1} simfrac{1}{2nsqrt n}$$ then refer to limit comparison test.
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
HINT
We have that for $n$ large
$$frac{sinleft(frac{1}{sqrt n }right)}{2n-1} simfrac{1}{2nsqrt n}$$ then refer to limit comparison test.
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
|
show 1 more comment
1
$begingroup$
HINT
We have that for $n$ large
$$frac{sinleft(frac{1}{sqrt n }right)}{2n-1} simfrac{1}{2nsqrt n}$$ then refer to limit comparison test.
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
|
show 1 more comment
1
1
1
$begingroup$
HINT
We have that for $n$ large
$$frac{sinleft(frac{1}{sqrt n }right)}{2n-1} simfrac{1}{2nsqrt n}$$ then refer to limit comparison test.
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
$endgroup$
HINT
We have that for $n$ large
$$frac{sinleft(frac{1}{sqrt n }right)}{2n-1} simfrac{1}{2nsqrt n}$$ then refer to limit comparison test.
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
answered Dec 5 '18 at 23:28
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
|
show 1 more comment
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
So is my approach absolutely wrong or is the whole substitution thing viable? Either in this problem or in other problems?
$endgroup$
– Eyad H.
Dec 5 '18 at 23:30
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. By comparison test we can use of course that $sin x <x$ for $x>0$ and then $$frac{sin(frac{1}{sqrt(n)})}{2n-1}le frac{1}{(2n-1)sqrt n}le frac{1}{2nsqrt n}$$ and conclude in the same way.
$endgroup$
– gimusi
Dec 5 '18 at 23:31
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
$begingroup$
@EyadH. The advantage by LCT is that we don't need to worry about inequalities which are more elegent but also much more subtle to handle or to remember!
$endgroup$
– gimusi
Dec 5 '18 at 23:34
1
1
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
@EyadH. I don't understand that step $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ I think that it is better to act directly with the original expression letting $n to infty$ to avoid confusion.
$endgroup$
– gimusi
Dec 5 '18 at 23:37
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
$begingroup$
I thought that since $-1 leq sin(u) leq 1$ and $frac{2}{u^2}-1 leq frac{1}{u^2}$ then it implied $frac{sin(u)}{frac{2}{u^2}-1} leq frac{1}{u^2}$ too. I don't know, I was pretty desperate to be honest and I was trying to hack myself through that ugly series and that seemed to make it easier and made sense.
$endgroup$
– Eyad H.
Dec 5 '18 at 23:52
|
show 1 more comment
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$begingroup$
Are you trying to figure out the sum of the infinite series, or do you just want to know if it converges/diverges?
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– Decaf-Math
Dec 5 '18 at 23:31
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I want to know if it's convergent or divergent AND know whether or not that u-substitution hack is valid.
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– Eyad H.
Dec 5 '18 at 23:36
1
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@EyadH. We should have $$frac{sin(u)}{frac{2}{u^2}-1} leq frac{u}{frac{2}{u^2}-1}= frac{u^3}{2-u^2}sim frac12 u^3=frac{1}{2nsqrt n}$$ which is exactly the same result we can obtain manipulating the original expression.
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– gimusi
Dec 5 '18 at 23:39
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@Decaf-Math I had the same doubt initially but it is clear from the context that the asker was looking for convergence/divergence.
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– gimusi
Dec 5 '18 at 23:41
1
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@EyadH. It is valid but at the end we need to reconvert to $n$. I think sincerly that it is not necessary, you can handle the original expression in a easy way, we don't need substitution. Bye
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– gimusi
Dec 5 '18 at 23:49