Mixing
Decide if the next infinite series are convergent or divergent.
$begingroup$
I need help with this problem:
Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.
- $sum_{n=1}^infty {{sin (ntheta)}over{n^2}}$
- $sum_{n=1}^infty (-1)^n {{log n}over{n}}$
I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks.
calculus sequences-and-series
edited Jan 28 at 4:28
Naman Kumar
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.
- $sum_{n=1}^infty {{sin (ntheta)}over{n^2}}$
- $sum_{n=1}^infty (-1)^n {{log n}over{n}}$
I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks.
calculus sequences-and-series
edited Jan 28 at 4:28
Naman Kumar
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.
- $sum_{n=1}^infty {{sin (ntheta)}over{n^2}}$
- $sum_{n=1}^infty (-1)^n {{log n}over{n}}$
I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks.
calculus sequences-and-series
edited Jan 28 at 4:28
Naman Kumar
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
$endgroup$
I need help with this problem:
Decide if the next infinite series are convergent or divergent. The tools that you need to use are the Leibniz Theorem, the Comparison Test, Quotient Test and the Integral Test.
- $sum_{n=1}^infty {{sin (ntheta)}over{n^2}}$
- $sum_{n=1}^infty (-1)^n {{log n}over{n}}$
I don't know how to start. Can you please help me and explain me how to decide when to use one of the tests? Thanks.
calculus sequences-and-series
calculus sequences-and-series
edited Jan 28 at 4:28
Naman Kumar
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
edited Jan 28 at 4:28
Naman Kumar
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
edited Jan 28 at 4:28
Naman Kumar
23013
edited Jan 28 at 4:28
Naman Kumar
23013
edited Jan 28 at 4:28
Naman Kumar
23013
23013
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
asked Jan 28 at 4:18
davidllerenavdavidllerenav
3078
3078
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$1$. Note that, Every absolutely convergent series is convergent.
Now, $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$. So, the series $sum_{n=1}^inftyfrac{sin (ntheta)}{n}$ is absolutely convergent and hence convergent.
$2$.For the second one we use alternating series test. Set $a_n=(-1)^nfrac{log n}{n}$ then, $|a_n|=frac{log n}{n}$ decreases monotonically as $n>log n$ for all $n>1$ with $displaystylelim_{nto infty}a_n=0$. Hence the series $sum_{n=1}^infty (-1)^nfrac{log n}{n}$ converges.
EDIT: Alternating series test, Absolute convergence, value of $sum_{n=1}^infty frac{1}{n^2}$
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
|
show 8 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$1$. Note that, Every absolutely convergent series is convergent.
Now, $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$. So, the series $sum_{n=1}^inftyfrac{sin (ntheta)}{n}$ is absolutely convergent and hence convergent.
$2$.For the second one we use alternating series test. Set $a_n=(-1)^nfrac{log n}{n}$ then, $|a_n|=frac{log n}{n}$ decreases monotonically as $n>log n$ for all $n>1$ with $displaystylelim_{nto infty}a_n=0$. Hence the series $sum_{n=1}^infty (-1)^nfrac{log n}{n}$ converges.
EDIT: Alternating series test, Absolute convergence, value of $sum_{n=1}^infty frac{1}{n^2}$
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
|
show 8 more comments
$begingroup$
$1$. Note that, Every absolutely convergent series is convergent.
Now, $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$. So, the series $sum_{n=1}^inftyfrac{sin (ntheta)}{n}$ is absolutely convergent and hence convergent.
$2$.For the second one we use alternating series test. Set $a_n=(-1)^nfrac{log n}{n}$ then, $|a_n|=frac{log n}{n}$ decreases monotonically as $n>log n$ for all $n>1$ with $displaystylelim_{nto infty}a_n=0$. Hence the series $sum_{n=1}^infty (-1)^nfrac{log n}{n}$ converges.
EDIT: Alternating series test, Absolute convergence, value of $sum_{n=1}^infty frac{1}{n^2}$
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
|
show 8 more comments
$begingroup$
$1$. Note that, Every absolutely convergent series is convergent.
Now, $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$. So, the series $sum_{n=1}^inftyfrac{sin (ntheta)}{n}$ is absolutely convergent and hence convergent.
$2$.For the second one we use alternating series test. Set $a_n=(-1)^nfrac{log n}{n}$ then, $|a_n|=frac{log n}{n}$ decreases monotonically as $n>log n$ for all $n>1$ with $displaystylelim_{nto infty}a_n=0$. Hence the series $sum_{n=1}^infty (-1)^nfrac{log n}{n}$ converges.
EDIT: Alternating series test, Absolute convergence, value of $sum_{n=1}^infty frac{1}{n^2}$
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
$endgroup$
$1$. Note that, Every absolutely convergent series is convergent.
Now, $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$. So, the series $sum_{n=1}^inftyfrac{sin (ntheta)}{n}$ is absolutely convergent and hence convergent.
$2$.For the second one we use alternating series test. Set $a_n=(-1)^nfrac{log n}{n}$ then, $|a_n|=frac{log n}{n}$ decreases monotonically as $n>log n$ for all $n>1$ with $displaystylelim_{nto infty}a_n=0$. Hence the series $sum_{n=1}^infty (-1)^nfrac{log n}{n}$ converges.
EDIT: Alternating series test, Absolute convergence, value of $sum_{n=1}^infty frac{1}{n^2}$
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
answered Jan 28 at 4:31
Sujit BhattacharyyaSujit Bhattacharyya
1,580519
1,580519
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
|
show 8 more comments
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
How did you found that $displaystylesum_{n=1}^infty |frac{sin(ntheta)}{n^2}|lesum_{n=1}^infty frac{1}{n^2}$? Why did you choose $sum_{n=1}^infty frac{1}{n^2}$? Sorry I find this topic pretty difficult.
$endgroup$
– davidllerenav
Jan 28 at 4:48
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Oh, I see. Recall that $|sin x|le 1$ for all $xinmathbb{R}$ so each term of the $n$-th partial sum of the above series i.e. $displaystylesum_{n=1}^N |frac{sin(ntheta)}{n^2}|$ is less than $sum_{n=1}^N frac{1}{n^2}$ for all $N=1,2,...$
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:51
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
$begingroup$
Or, simply $|frac{sin(ntheta)}{n^2}|lefrac{1}{n^2}$ and take finite sum both sides then take the sum to infinite sum.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 4:53
1
1
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
So $|sin(ntheta)|$ is always $leq 1$? And then the inequality will remain when we divide by $n^2$, right?
$endgroup$
– davidllerenav
Jan 28 at 5:00
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
$begingroup$
absolutely correct. and it is true for all n and for all theta.
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 5:02
|
show 8 more comments
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