Evaluate $$\sum_{n=0}^\infty (n+1)\left(\frac13\right)^n.$$   $$1 + 2\left(\frac{1}{3}\right) + 3\left(\frac{1}{3}\right)^2 + 4\left(\frac{1}{3}\right)^3 + 5\left(\frac{1}{3}\right)^4 +\cdots$$  Show using ‘Techniques of Convergence of Geometric Series’ only that it converges to 9/4?
1602
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$
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. $$ \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} $$
34345
I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$  I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:  For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.
59349
I am given the following series:  $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$$   I have used the alternating series test to show that the series converges.   However, how do I go about showing what it converges to?
9865
I think the radius of convergence for  $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$, $x\in \mathbb R$ is:   $r^{-1}$=$\lim_{n\to \infty}$$|\frac{a_{n+1}}{a_n}$|=1   so we get that $r$=1.  But how can I show it formally?  After that, I have to show that $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$ is uniformly convergent in the interval $[-r,r]$. Can I maybe use Weierstrass majoranttest? What can I use as majorant serie?
34609
how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $? It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...  Let consider sum no to infinity, but to n  $$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$ $$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$ but $$ n = ln e^n $$ so it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$  So the limit of it is $-\infty$ Have I done this well or I missed sth?
123432
I have the following series  $$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$  I have found out that the series is equal to  $$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$  I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.  Does anyone have any idea how I can go about doing this ?
28686
Finding $$\sum^{\infty}_{n=1}\bigg[\bigg(\frac{(-1)^{n+1}}{n+1}\bigg)\bigg(\sum^{n}_{r=1}\frac{1}{r}\bigg)\bigg]$$   Try: Let $\displaystyle H_{n} =\sum^{n}_{r=1}\frac{1}{r},$ then series $\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n+1}\cdot H_{n}=\frac{H_{1}}{2}-\frac{H_{2}}{3}+\frac{H_{3}}{4}\cdots\cdots$  could some help me to solve it, thanks
3264
Evaluate : $$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$$ I tried using logarithmic functions, but cannot evaluate this please help me
63056
Does the convergent infinite sum $$ \sum_{n=0}^{\infty} \frac{1}{2^n + 1} $$ have a closed form solution? Quickly coding this up, the decimal approximation appears to be $1.26449978\ldots$
45441
I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?  HERE IS ONE   Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then, \begin{align} a_n&amp;=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&amp;=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&amp;=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&amp;=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&amp;=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align} Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that  \begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align} Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$ \begin{align} \exp \left(-\sqrt{n}\right)&lt;\dfrac{1}{n^2}.\end{align} Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}&lt;\infty,\end{align} and so, $\sum^{\infty}_{n=1}b_n&lt;\infty\implies \sum^{\infty}_{n=1}a_n&lt;\infty$ by Limit comparison test.
60070
How can we evaluate this series  $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$  Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?  Isn't there a simple solution which doesn't involve Stirling numbers?
68504
I was doing math and I came across this series     $$\sum_{n=1}^\infty\sin n\cdot\sin\left(\dfrac1n\right)$$   I managed to prove that this series is convergent, but I was wondering what value the series converges to $?$ I've put it on Scientific workplace and got a value of around $0.92 $ But I have no idea how to get there.
79381
I have to study the convergence of the series $$\sum_{n=1}^\infty \left(1-n\sin\left(\frac{1}{n}\right)\right).$$ I know I should study the limit $$\lim_{n \to \infty}n^\alpha\left( 1-n\sin\left(\frac{1}{n}\right)\right).$$   But I have some troubles with this limit (and so the convergence of the series). Any ideas?
30910
Could I have a hint for testing the convergence of the following series please?  $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$  I am very appreciative for your help.
2301
I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done?
113494
Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..