I know the series $$\sum_{k=1}^\infty \frac{(\frac{1}{10})^{k+1}}{k(k+1)}$$ is equal to $$\frac{1}{10}-\frac{9}{10}\ln \Big(\frac{10}{9}\Big)$$ but how would I use the ratio test to show this is true?  I also know that:  $$ (1-x)\ln(1-x)=\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=1}\frac{x^k}{k} =\sum_{k=1}^\infty \frac{x^{k+1}}{k(k+1)}$$
122884
Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.   This is my idea.  We have the radius of convergence is $R=1$. And $\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.  Then let $f(x)=\sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.  We have $f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ and  $f''(x)=\sum_{n=1}^{\infty} x^{n-1}=\dfrac{1-x^n}{1-x}$  Then if I take the primity of $f''(x)$, I get: $f'(x)=\displaystyle\int\limits_{0}^{x}\dfrac{1-t^n}{1-t}\mathrm{d}t=-\ln(1-x)-\displaystyle\int\limits_{0}^{x}\dfrac{t^n}{1-t}\mathrm{d}t$  Then I will get a sum again, and that's not the goal.  What can I do then? Thank you so much.
29480
I would like to check if the series $\sum_{m=0}^{\infty}  {2m\choose m} \frac{1}{4^{m}} $ converges or diverges. The ratio test is inconclusive. Any other way to proceed? Thanks.
52305
I was doing some problems in algebraic number theory and this series came up  $$\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}.$$  So, I would like to know the value of this series. However, I don't want a full answer, only hints. I have tried to use Fourier's series, but with no success, perhaps an answer can be gotten usign this.  Thanks
34891
I need to find out whether the following series converges or diverges. If it converges I also have to discover whether it is an absolute convergence or a conditional one. $$\sum_{k=1}^{\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^{2k}$$ Root test did not help me much since $\lim_{k\rightarrow\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^2$ may equal anything from $0$ to $4$. Comparison tests also did not work well. I guess this series is convergent, though. But I still have no idea how to prove it.
41159
Does $$\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}$$ converge? If yes, to where?  I searched this specific series but couldn't find a solution.
36066
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test. $$ \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} $$
51854
I'm having trouble determining the convergence of the series:  $$ \sum_{n=1}^{\infty}\left[1-\cos\left(1 \over n\right)\right]. $$  I have tried the root test:  $$\lim_{n\rightarrow\infty}\sqrt[n]{1-\cos\frac{1}{n}}=\lim_{n\rightarrow\infty}\left(1-\cos\frac{1}{n}\right)^{1/n}=\lim_{n\rightarrow\infty}\mathrm{e}^{\frac{\log(1-\cos\frac{1}{n})}{n}}=\mathrm{e}^{\lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n})}{n}}$$  Now by applying the , that upper limit is equal to:  \begin{align} \lim_{n\rightarrow\infty}\frac{\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})}{(n+1)-n}&amp;=\lim_{n\rightarrow\infty}\left(\log(1-\cos\frac{1}{n+1})-\log(1-\cos\frac{1}{n})\right) \\&amp;=\lim_{n\rightarrow\infty}\log{\frac{1-\cos{\frac{1}{n+1}}}{1-\cos{\frac{1}{n}}}} \end{align}  Now I'm totally stuck, unless that quotient is actually 1, in which case the limit would be 0, the Root test result would be $\mathrm{e}^0=1$ and all this would have been to no avail.  I'm not sure this method was the best idea, the series sure seems way simpler than that, so probably another method is more appropriate?
79495
I am looking at some homework where I have:  $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$  How can I sum this?   I know that   $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  and also that  $\sum_{n=1}^\infty \frac{1}{n(n+1)}=1$  But I just can't see or find the connection
22902
I need to prove that:  $$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$$  is convergent, but not absolutely convergent.  I tried the ratio test:  $$\frac{a_{n+1}}{a_n} = -\frac{\log\left(1+\frac{1}{n+1}\right)}{\log\left(1+\frac{1}{n}\right)} = -\log\left({\frac{1}{n+1}-\frac{1}{n}}\right)$$  I know that the thing inside the $\log$ converges to $1$, so $-\log$ converges to $0$? This is not right, I cannot conclude that this series is divergent.  Also, for the sequence without the $(-1)^{n+1}$ it would give $0$ too.
74045
Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$   so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since  both series are convergent so the sum of it is also convergent but I'm not sure if it's legall   second series I'm not sure is like this $\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$   here I also split it into smaller parts namely $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}$ and I prove that $\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}&lt;\frac{1}{n^2} \cdot c $ for some const c so it implies that origin series is convergent or not ?
38229
Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?  Thanks!
10024
I want to find the convergence interval of the infinite series $\sum\limits_{n=0}^\infty \dfrac{n!x^n}{n^n}$. I will use the ratio test: if I call $u_n = \dfrac{n!x^n}{n^n}$, the ratio test says that, if the following is true for some values of $x$, the series will be convergent for these values of $x$: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|&lt;1$$ So, I will first calculate the value of $\left|\dfrac{u_{n+1}}{u_n}\right|$: $$\left|\dfrac{\dfrac{(n+1)!x^{n+1}}{(n+1)^{n+1}}}{\dfrac{n!x^n}{n^n}}\right|=\dfrac{(n+1)!|x|^{n+1}}{(n+1)^{n+1}}\times\dfrac{n^n}{n!|x|^n}=\frac{(n+1)n^n|x|}{(n+1)^{n+1}}=|x|\left(\frac{n}{n+1}\right)^n$$ So, $\lim\limits_{n\to+\infty}\left|\dfrac{u_{n+1}}{u_n}\right|$ becomes: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim_{n\to+\infty}|x|\left(\frac{n}{n+1}\right)^n=|x|\lim_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$$ Now I must evaluate the value of $\lim\limits_{n\to+\infty}\left(\dfrac{n}{n+1}\right)^n$. For this, let $y = \left(\dfrac{n}{n+1}\right)^n$; so, instead of calculating $\lim\limits_{n\to+\infty}y$, I will first calculate $\lim\limits_{n\to+\infty}\ln y$: $$\lim_{n\to+\infty}\ln y=\lim_{n\to+\infty}\ln \left(\dfrac{n}{n+1}\right)^n=\lim_{n\to+\infty}n\ln\left(\frac{n}{n+1}\right) =\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}}$$ Applying L'Hôpital's rule: $$\lim_{n\to+\infty}\frac{\ln\left(\frac{n}{n+1}\right)}{\frac{1}{n}} =\lim_{n\to+\infty}\frac{\frac{1}{n(n+1)}}{-\frac{1}{n^2}} =\lim_{n\to+\infty}\left(-\frac{n}{n+1}\right)=-1$$ Now, since we know that $\lim\limits_{n\to+\infty}\ln y = -1$, we have that: $$\lim_{n\to+\infty}y=\lim_{n\to+\infty}e^{\ln y} = e^{-1} = \frac{1}{e}$$. Substituting this back into the expression $\lim\limits_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right| = |x|\lim\limits_{n\to+\infty}\left(\frac{n}{n+1}\right)^n$, we have that the limit of $\left|\dfrac{u_{n+1}}{u_n}\right|$ as $n\to+\infty$ is: $$\lim_{n\to+\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x|}{e}$$ Therefore, the series will certainly be convergent for the values of $x$ for which $\dfrac{|x|}{e}&lt;1$, that is, $|x|&lt;e$. So, I know that the series is convergent for $-e &lt; x &lt; e$, but I have to test whether the series is convergent at $x = e$ or $x = -e$. That is, I have to test whether $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ and $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ are convergent. Since these limits don't approach zero, I know they are both divergent, but I'm not sure how to find the limits, because of the factorial function. Also, I can't use integral test here, because of the factorial. Probably I should use comparison test, but I haven't found any divergent series to which to compare it. Any hints? Thank you in advance.  Edit: Using the suggestion by Ragib Zaman in the answer below, since the Taylor polynomial $P_n(x)$ of $e^x$ at $a=0$ is $$e^x = 1 + x + \dfrac{x^2}{2!} + \cdots + \dfrac{x^n}{n!}+\cdots,$$ if we substitute $n$ for $x$ we see that $e^n&gt;\dfrac{n^n}{n!}$; therefore, $\dfrac{n!e^n}{n^n} &gt; 1$, and, thus, we show that $\sum\limits_{n=0}^{\infty} \dfrac{n!e^n}{n^n}$ is divergent, because its term doesn't approach zero. $\sum\limits_{n=0}^{\infty} \dfrac{(-1)^nn!e^n}{n^n}$ is also divergent, because, although the absolute value of the ratio between two successive terms, $|e|\left(\frac{n}{n+1}\right)^n$, approaches 1 as $n\to\infty$, it approaches 1 from values bigger than 1; therefore, the absolute value of a term is always greater than the absolute value of the previous term.
113978
I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
82381
Is the following series convergent?  $$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$  I treid the Ratio and Root tests, but both of them failed.
3340
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 $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:     What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$
34945
I am sorry that my question is not whether $\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}$ convergent or not.  My question is how to compute the sum of series:  $$\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},$$ It is a question in a textbook, and I have worked on it for days, but fail.
80992
I tried to demonstrate the convergence of the following series by root and ratio test but they failed.  $$\sum_{n=1}^{\infty}\frac{\log(n)}{n^2}$$  Then I used Wolfram Alpha, which answered "By the comparison test, the series converges".  How can I do it?