Does $|x_k|$ $\to$ $0$ and $\sum_{k=1}^{\infty} |x_k| &lt; \infty$ $\implies$ $\sum_{k=m+1}^{\infty} |x_k| \to 0$ as $m\to\infty$ ? If so, how would one formally prove this? Any help would be greatly appreciated.
123238
Let $x_n$ be a sequence of real numbers.  Show that if $\sum_{n=1}^{\infty}x_n $ converges, then $\lim_{n\rightarrow\infty}\sum_{k=n}^{\infty}x_k=0.$  This is what I have done. We have by the Cauchy Criterion, for every $\epsilon&gt;0$ there is an $N\in\mathbb{N}$ such that $n\ge N \ \implies |\sum_{k=n}^{\infty}x_k|&lt;\epsilon.$ Taking the limit as $n \rightarrow \infty$ we get the desired result.  Are there any problems with my proof?
110774
Does this double infinite sum $\sum_{n=1}^\infty \sum_{m=1}^\infty\frac{1}{(m+n)^2}$ converge?  I can not understand how Ido this Please step by step
30633
Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?  The ratio test is inconclusive.  It passes the term test for $\Re(n)=1$, but this is not sufficient to prove convergence.  Since we are dealing with so many complex numbers, I do not know of any convergence tests for this.  I know that if it is convergent, it is conditionally convergent.
21881
Let $(x_n)$ a decreasing sequence and $\sum x_n\to s$. Then $(n\cdot x_n)\to 0$   Check my proof please, Im not completely sure about it correctness.  If $\sum_{k=h}^\infty x_k= s$ then we can rewrite the sum for starting index $1$ with the change $k-h=j$, then  $$\left(\sum_{j=1}^n x_j\right)-n\cdot x_n=\sum_{j=1}^n (x_j-x_n)$$  Then taking limits  $$\color{red}{\lim_{n\to\infty}\left[\left(\sum_{j=1}^n x_j\right)-n\cdot x_n\right]}=\lim_{n\to\infty}\sum_{j=1}^n (x_j-x_n)=\sum_{j=1}^\infty (x_j-0)=\color{red}{\lim_{n\to\infty}\sum_{j=1}^n x_j}=s$$  where I used the fact that $(x_n)\to 0$. Thus equating the colored expressions this implies that $\lim_{n\to\infty} nx_n=0$.    The proof, to my eyes, seems correct but I dont needed in any moment to use the fact that $(x_n)$ is a monotonic sequence so it is possible that I make a mistake somewhere or that the proof is incorrect.    My second attempt  Because $\sum x_k$ converges and is positive (cause $(x_n)\downarrow 0$) we can write   $$\sum_{k=n+1}^{2n+m}x_k=\left|\sum_{k=n+1}^{2n+m}x_k\right|&lt;\epsilon/2,\quad \forall n,m\ge N$$  Then, cause $(x_n)$ is decreasing  $$(n+m)x_{2n+m}\le\sum_{k=n+1}^{2n+m}x_k&lt;\epsilon/2\\\implies (2n+m)x_{2n+m}\le2(n+m)x_{2n+m}&lt;\epsilon,\quad\forall n,m\ge N$$  Because $m$ is arbitrary setting $M=2N&gt;N$ we can finally write  $$nx_n&lt;\epsilon,\quad\forall n\ge M$$  It is this proof correct? Thank you.
45031
Exericise 9.36: If each $a_n &gt;0$ and $\sum a_n$ diverges, prove that $\sum a_n x^n \to +\infty$ as $x\to1^-$. (Assuming $\sum a_n x^n$ converges for $|x|&lt;1$.   Here's what I tried:  Since $\sum a_n$ diverges, for $M&gt;0$, $\exists N$ such that $\sum_{k=0}^N a_k &gt;M.$ I want to show that for $M&gt;0$, there exists $\delta$ such that, whenever $\delta&lt;x&lt;1$ implies that there exists $p$ such that $\sum_{k=0}^p a_k&gt;M$.  I was reading  and I don't get how the "then by continuity" part implies (**) part. Could someone explain? Thanks!
44043
We let a>0 and b>0 and  we have $\sum_{n=0}^{\infty}b^n-a^n$. We have to find all values a,b>0 where $\sum_{n=0}^{\infty}b^n-a^n$ is convergent. I think I can use that $$\sum_{n=0}^{\infty} b_n &lt; \infty , \sum_{n=0}^{\infty} a_n &lt; \infty \implies \sum_{n=0}^{\infty} (b_n-a_n) &lt; \infty. $$. And I think $\sum_{n=0}^{\infty}b^n$ and $\sum_{n=0}^{\infty}a^n$ is convergent if a&lt;1 and b&lt;1? It will make sense. But how can I show it?
21705
I got this sum, in some work related to another :  $$S_m=\sum_{k=1}^m  \frac{1}{k}{m \choose k} $$  Are there any known results about this (bounds, asymptotics)?
38666
$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?
7191
Discuss the convergence of the sum:   $$\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}.$$  My answer so far: It does not converge absolutely since   $$\left | \frac{(-1)^k}{\sqrt{k}+(-1)^k} \right | \geq \frac {1} {\sqrt{k} + 1},$$  the sum of whose terms diverges. Intuitively, I think it should converge conditionally because $ \sum \frac{(-1)^k}{\sqrt{k}}$ does, and the $(-1)^k$ in the denominator should not make a big difference. However, I am having trouble demonstrating its conditional convergence.  Evidently, the problem is that $\frac{1}{\sqrt{k}+(-1)^k}$ is not monotone. Let's call this sequence $a_n$. If we can show that   $$\sum_{k=2}^\infty |a_k - a_{k-1}| &lt; \infty,$$   then we're done by Dirichlet's test. But this seems to be $O(1/k)$.   Any ideas?
122146
Exercise: Suppose that $a_k \geq 0$ for  $k$ large and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges. Prove that $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} = 0$$  Attempt in proof: Suppose  $a_k \geq 0$ for any large $k$ and that $\sum_{k = 1}^{\infty} \frac{a_k}{k}$ converges. . Then give $\epsilon &gt;0$  there is $N \in N$ such that $\left|\sum_{k = n}^{\infty} \frac{a_k}{k}\right| &lt; \epsilon$. Let $s_n = \frac{a_n}{j+k}$ denote the partial sums. Then $\sum_{k = 1}^{\infty} \frac{a_k}{j+k} $ will converge to zero if and only if its partial sum converges to zero as n approaches infinity. Taking the limit of $$\lim_{j \to \infty}\sum_{k = 1}^{\infty} \frac{a_k}{j+k} =  \lim_{j \to \infty}\frac{a_1}{j + k} + \cdots+ \frac{a_n}{j + k} + \cdots = \lim_{j \to \infty}\frac{a_1/j}{1 + k/j} + \cdots+ \frac{a_n/j}{1 + k/j} +\cdots $$ Can someone please help me finish? I don't know if this a right way. Any help/hint/suggestion will be really appreciate it. Thank you in advance.
62940
Assume that $0 &lt; a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that  $$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} &lt; \infty $$ ?  I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon &gt; 0$ this is obvious.  Any idea? Thanks.
68274
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$  in $n$?  I know this is well-known. But how to prove it rigorously? Even mathematical induction does not seem so straight-forward.  Thanks.
114562
Decide if the series converges and prove it using comparison test:   $\sum_{k=1}^{\infty}\frac{1}{2k+3}$   $$\frac{1}{2k+3} &lt;\frac{1}{2k}&lt;\frac{1}{k}$$  and since $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too.  Did I do everything correctly? What makes me feel a bit unsure is that there is no "$\leq$" sign anywhere and when I look up the comparison test on the internet, I don't see any "&lt;" or ">" in the definition.  Is it correct anyway?
47290
This is the other direction of  that reads  Let $a_n&gt;0$ (!) for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty a_n&lt;\infty$ implies $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}&lt;\infty$.  I need to prove that &quot;the converse of the last statement is false&quot;. I don't want the answer, but to ask if I can use the following argument to prove the statement. To start, I'm confused with the statement &quot;the converse of the last statement is false&quot;. Is it enough if I prove, if $\sum_n \sqrt{a_n a_{n+1}}$ diverges, then $\sum_n a_n$ must diverge? If this is true, then the proof is easy since I can choose $a_n=1/n$ and prove that $\sum_n a_n$ converges.
65491
Let $(x_n)_{n=1}^{\infty}$ be a sequence such that $|x_{n+1} - x_n| &lt; r^n$ for all $n \geq 1$, for some $0 &lt; r &lt; 1$. Prove that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.  I understand that a Cauchy sequence means that for all $\varepsilon &gt; 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. But this one is really giving me a headache.  I tried doing something like: let $ m &gt; n$. Therefore $x_n - x_m$ =  $(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) $ and then somehow using the triangle inequality to compute some sum such that $x_n - x_m$  &lt; sum which would be epsilon?  any help is appreciated, thank you.
117373
Any hint to prove this? I tried with properties of binomial coefficient but I can't get anything. $$\sum_{n=k}^{\infty}{{n}\choose{k}}x^n=\dfrac{x^k}{(1-x)^{k+1}}$$
20776
Question : Is the following true for any $m\in\mathbb N$?   $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$   Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?  By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.  Proof : Let  $$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$ We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get $$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$ Then, the sum of these from $1$ to $n$ satisfies  $$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$ Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.