Prove that, if $a_k \geq 0$ for large $k$, and that $$\sum_{k=1}^\infty \frac{a_k}{k}$$ converges. Prove that $$\lim_{j \rightarrow \infty} \sum_{k=1}^\infty \frac{a_k}{j+k} = 0$$  I believe I cant write  $$\sum_{k=1}^\infty \frac{a_k}{j+k} \leq \sum_{k=1}^\infty \frac{a_k}{k}.$$  However, taking limit does not give me the answer.
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.
52746
Does the condition $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?  When I say convergent I mean to $L\in \mathbb{R}$ or to $\infty$ or to $-\infty$  I have a gut feeling this statement is false, just can't find a series which contradicts it.  Thanks a lot
99641
I'm looking for slick proofs that if $a_n$ is a sequence of complex numbers such that $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges then $\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0$.   My not so slick proof:      Let $A_x=\sum\limits_{k=1}^{x} a_k$ and apply an Abel sum with the function $f(x)=\frac{1}{x}$. We get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\int\limits_1^n\frac{A_x}{x^2}dx=\frac{A_n}{n}.$$ Of course we can split the integral into chunks to get $$\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{i=1}^n \left( \frac{1}{k+1}-\frac{1}{k} \right) A_k=\frac{A_n}{n}.$$ If we expand $A_k$ on the left side, it telescopes and yields $$\sum\limits_{k=1}^n \frac{a_k}{k} + \sum\limits_{k=1}^n \left( 1-\frac{1}{k+1}\right) a_k=\frac{A_n}{n}\iff \frac{(n-1)A_n}{n}=\sum\limits_{k=1}^n\frac{a_k}{k(k+1)}.$$ The norm of the right side  of the equation is clearly $\mathcal O(\log(n))$ (using the triangle inequality and the fact that $\frac{a_k}{k}$ is bounded). The desired result follows after dividing both sides by $n-1$.   I would also appreciate some proof verification. Initially, I thought that the problem was going to be really easy, but it took me a bit of effort. Am I missing something major? The original problem is for real sequences, but I don't think this helps. Obviously if the sequence converged absolutely then it would also be absolutely trivial.
95604
Can anyone help me with the following question.   If $a_{n} \geq 0$ and $\sum a_{n}$ converges then how to prove $\sum \frac{\sqrt{a_{n}}}{n}$ converges.    Any idea where to start. My idea was to try using comparison test since $\sqrt{a_{n}} \leq a_{n}$ but it appears that it wouldn't work if $0 \leq a_{n} &lt;1$.
48581
Firstly we are given that $a_k\in\mathbb C$ and $\sum\limits_{k=1}^\infty {a_k}^2$ converges absolutely. The question then asks to show that $\sum\limits_{k=1}^\infty \frac{a_k}{k}$ converges absolutely.  I know that absolute convergence of the first series this implies that $\sum\limits_{k=1}^\infty ||{a_k}^2||$ converges.  Thus by the Cauchy-Schwarz inequality I get  $$ \sum\limits_{k=1}^\infty ||{a_k}^2|| \leq \sum\limits_{k=1}^\infty ||{a_k}||.||a_k|| $$  From this point onwards I have no idea of how to proceed. (Should I make use of Cauchy products?)
39500
Let $$ a_{n+1} =  \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k&gt;1, a_1&gt;0 $$ The problem is to show that it converges.  Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track.  Please provide any hints.
23840
I just completed the following proof.  Is it valid?   Let $\sum_{k=1}^{\infty} a_k$ be an arbitrary convergent series that also converges absolutely.  Then $\sum_{k=1}^{\infty} a_k \in \mathbb{C}$ and $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$. Let $\{b_k\}$ denote a rearrangement of the $\{a_k\}$.  That is, we have that $\{a_k\} = \{b_k\}$ yet $a_i = b_i$ needn't hold. We will first show that $\sum_{k=1}^{\infty} b_k$ converges absolutely (immediately implying it converges non-absolutely).  We will use the Cauchy Criterion on the partial sums of $\sum_{k=1}^{\infty} |b_k|$ to demonstrate this. First let $\epsilon &gt; 0$. Consider that since $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$, we have that $\exists N_1 \in \mathbb{N}$ s.t. for all $N_1 &lt; m &lt; n$ we have   $$ \sum_{k=m}^n |a_k| \le \epsilon. $$ Now let $0 &lt; N_2$ be large enough s.t $\{a_1, \ldots , a_{N_1}\} \subseteq \{b_1, \ldots , b_{N_2}\}$.  Let $M &gt; max\{N_1, N_2\}$.  Let $M &lt; m' &lt; n'$.  Then since $\{b_k\} - \{b_1, \ldots , b_{m'}\} \subseteq \{a_{N_1}, a_{N_{1}+1},  a_{N_{1}+2} \ldots\}$, we will have that   $$ \sum_{k=m}^n |a_k| \le \epsilon \implies \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon. $$ Then we have that   $$ \left| \sum_{k={m'}}^{n'} |b_k| \right| = \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon $$  so that since $\epsilon$ was arbitrary, we have obtained that   $$ \left| \sum_{k=1}^n |b_k| - \sum_{k=1}^m |b_k|  \right| = \left| \sum_{k=m}^n |b_k|  \right| \rightarrow 0 $$  so that via the Cauchy Criterion for Convergence  we have that $\sum_{k=1}^{\infty} |b_k|$ converges to some  value in $\mathbb{R}$ as desired. Then it follows that $\sum_{k=1}^{\infty} b_k$ is absolutely convergent (and hence also just convergent).   $\square$
123411
I was wondering how the convergence of the series $$\sum_{k\geq1}\frac{(k+1)^y-k^y}{k^x},$$ depends on $x$ and $y$, where $x$ and $y$ are real numbers. I've been able to show that it converges if and only if $y\leq x$, by distinguishing nine cases; whether $x$ and $x-y$ are positive, negative or zero. If desired I can include a sketch of my proof. My question is whether there is a proof that works for all real $x$ and $y$? Or at least a proof that seems less artificial?
44779
Please help me with this question. I've been thinking about it for almost two days. Let $a_n$ a real series that have the following property: for every series $b_n$ in $l_2$: $\sum_{n=1}^\infty a_nb_n$ converges. prove that $a_n$ in $l_2$.
43842
Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$ Prove that the series converges (resp. diverges) if $L &gt; 1$ (resp. $L&lt;1$). I've tried, for example, that when $L&lt;1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?  P.S. using Kummer's test doesn't count
92775
Let $\alpha$ be a positive number. Calculate $$\lim_{n\to\infty} \sum_{k=1}^{n} \Big(\frac{k}{n}\Big)^{\alpha k}$$  Edit: I have deleted my attempt, it didn't seem to lead me anywhere and I discovered a couple of mistakes. :)  Please give me a hint!
76156
I need help with the following limit $$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{\sqrt{kn}}$$  Thanks.
93786
Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.   I see  that,  $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} &lt; \sum_{k=1}^{n} \frac{1}{n} = 1 $.   Thank you.
56635
I am not able to solve this exercise. I've tried everything.     Exercise Let be $a_n &gt; 0$ a sequence and suppose that $\sum a_n$ converges. Prove that exists a sequence $c_n &gt; 0$ such that $\lim c_n = \infty$ and $\sum c_na_n &lt; \infty$   I don't even know if I have to find an "always-valid" closed form for $c_n$ (maybe in function of $a_n$?) or prove its existence in another way.  Thanks in advance.
23237
Let $(a_n)_n$ be a sequence of non-negative real numbers.      Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.   I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.  Could any convergence test like the ratio or the root test help?
119835
I'm trying to prove the convergence of $$ \sum_{n=1}^{\infty}\frac{1}{n^\alpha}$$ with $\alpha &gt; 1$.   For $\alpha \geq 2$ I can use the comparison test ($\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges) so I'm missing $2&gt;\alpha&gt;1$ and I'm pretty much out of ideas.  If you could offer some advice I would very much appreciate it. Thanks in advance
60987
We had the following theorem in class:     Let $(a_n)$ and $(b_n)$ be sequences and $b_n&gt;0$ and $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=L$ with $L\in\mathbb R\backslash\{0\}$. Then $\sum a_n$ converges if and only if $\sum b_n$ converges.   So by recapitulating the lecture I've tried to prove it but I didn't get it. It's obvious that you have to use the comparison test, but how? Can anybody help? Thanks a lot!