The question is basically to prove that convergence of $\sum{a_n}$  implies convergence of $\sum {\frac{\sqrt a_n}n}$  with $a_n&gt;0.$  Basically what I tried to show is that:  Since $\sum{a_n}$ converges, then $\lim a_n= 0$ when $n \rightarrow \infty$ , and so $\sqrt a_n \rightarrow 1$ when $n \rightarrow \infty,$ which means it is bounded. On the other hand, $\frac1n \rightarrow 0$ as $n \rightarrow \infty$ and is always decreasing. Per theorem 3.42 in Rudin, that means $\sum{\frac{\sqrt a_n}n}$ also converges.  Is that correct or am I missing something? The solutions on the net appear to differ.
114979
This is Baby Rudin exercise 7 of Chapter 3. Prove that the convergence of $\sum{a_n}$ implies the convergence of $\sum{\sqrt {a_k} \over k}$ if $a_n \ge 0$.  Proof:  I will attempt to show that the cauchy criterion is met so $${\sum_{k=n}^m {\sqrt {a_k} \over k}} \le \epsilon $$ for all $n,m \le N_1$. Im disregarding the absolute value since its all positive terms.   Since $\sum{a_n}$ converges, we know that the $\lim \limits_{n \to \infty} a_n =0$ So there exists some $N_2$ such that for all $n \ge N_2$, $$   a_n \lt \ {{(m-n)}^2k^2(\epsilon)^2} $$ I am again disregarding the absolute values because its always positive Let N=max{$N_1,N_2$} then for all $n,m \ge N$ $$\sum_{k=n}^m {\sqrt {a_k} \over k} \le \sum_{k=n}^m {\sqrt {{(m-n)}^2k^2(\epsilon)^2} \over k} \le \sum_{k=n}^m{(m-n)(\epsilon)} = \epsilon $$  Maybe this is just crazy, should I just use Cauchy-Schwarz inequality and forget about this or could I possibly make this method work?
15457
Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
83494
If $\sum_{n=2}^{\infty}a_n$ converge, then also converge this series? $$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)$$  Please verify my answer below  Counterexample:  $$a_{n}=\begin{cases} \frac{1}{k^{2}} &amp; n=k!^{k^{2}}\\ \\ 0 &amp; \text{All other cases} \end{cases}$$  When our infinites sum is equal to  $$\sum_{k=1}^{\infty}\frac{1}{k}\cdot\frac{1}{2\, \ln\, k}\cdot(k!-1)$$
61842
Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.   This is in the section on the Comparison Test so that must be what I'm supposed to use.  But I don't see how.  $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$.  And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges.  Hmm.  Any hints?
45816
For $a_n \geq 0$, and $\sum a_n$ convergent, show that  $\sum \frac{\sqrt{a_n}}{n^p}$ is also convergent for $p &gt; 1/2$?  What bugs me more is why isn't $\sum \sqrt{\frac{a_n}{n}}$ convergent??  Clearly it converges for $p &gt; 1$, and $a_n$ somehow helps out in $p \in (1/2,1)$.  Only hints are welcomed  Thanks in advance..
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?
70086
I'm stuck on a homework question, and could really use some help. Here is said question:  "Assume that for every $n$ the following occurs: $|a_{n+1}-a_{n}|&lt;q|a_{n}-a_{n-1}|$ when $ 0&lt;q&lt;1 $  Prove that the series $ \left(a_{n}\right)_{n=1}^{\infty} $ converges. Hint: use Cauchy sequence"  OK I'm really stuck on this one, I don't even have a clue where to begin.  Any help is appreciated. Thank you!
112564
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
30657
Following the guidelines suggested in , I am going to post a proposed proof as an answer to the theorem below. I believe the proof works, but would appreciate any needed corrections.  Theorem If a series $\sum_{n=1}^{\infty}a_n$ of real numbers converges then $\lim_{n \to \infty}a_n = 0$
69560
Prove that if $a_n\gt 0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ diverges.     This is the solution to this problem, but I'm having a hard time understanding it. Why does $a_k/(1+a_k)$ not converge to $0$ if $a_k$ doesn't converge to $0$?   I'd appreciate it if anyone could answer this question for me.
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.
26570
The last problem I posted had a wrong statement. I recovered the correct one.  Let $(a_n)$ be a sequence of positive real numbers. Prove that $\sum_{n\geq 0}a_n$ converges iff $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ converges  The direct statement is easy to prove using comparison test. I'm (again!) stuck with the converse. I tried summation by part, without success. Note that the convergence of $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ implies $\displaystyle \frac{a_n}{\sum_{k=0}^n a_k} \to 0$ I don't know what to do next ...
53026
$\{a_n\}$ sequence $a_1=\sqrt{6}$ for $n \geq 1$ and $a_{n+1}=\sqrt{6+a_n}$ show that it convergence and as well find $\lim \limits_{n \to \infty} a_n$  In order to show that that sequence convergence I need to show that :  $$\lim_{n \to \infty} a_n= L$$  While $L$ is finite.  Using the calculator. I assume that L=3 because :  $$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6........}}}}=2.999 \cong  3$$  I really don't think that this method is good enough to established that $\lim \limits_{n \to \infty} a_n= 3$ since it based on intuition.  I'll be glad to hear any ideas for an established method to show this.?  Any help will be appreciated.
46234
Let ${a_n}$ and ${b_n}$ sequences with positive terms, such that $\sum a_n$ and $\sum b_n$ converges. Prove that $\sum a_nb_n$ converges as well.  What I did: $\sum a_nb_n \le \left(\sum a_n\right) \left(\sum b_n\right)$ and by comparison test we are done.  However, when I asked my teacher regarding that solution, he told me that I can assume that inequality for finite numbers but for infinite numbers, it is not valid. It seems to me that $\left(\sum a_n\right) \left(\sum b_n\right)$ is absolute converges so that are no Riemann issues with this.  Can any one enlighten me and explain why is this invalid?
79337
Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.  My attempt: If  $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|&lt;|\sqrt2-a_{n-1}|$ but couldn't.
39681
If $\sum\limits_{n=1}^\infty a_n$ converges where $a_n&gt;0,\ \forall n\in\mathcal N$, prove $\sum\limits_{n=1}^\infty \sqrt[n]{a_1a_2\cdots a_n}$ and $\sum\limits_{n=1}^\infty\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$ converge.  $\sum\limits_{n=1}^\infty \frac{\sum\limits_{i=1}^na_i}{n}$ diverges, so one can not use the bound of the GM-AM relation at least directly.
61480
We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.  I want to prove that $a_n$ converges. But first I'd like to show that $a_n&lt;\sqrt{2}$ for every $n\in\mathbb{N}$.  I try to use induction. For a fixed $n$, I suppose that $a_n&lt;\sqrt{2}$.  Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)&lt;\sqrt{2}$?  Thanks.