If $\sum u_n^2$ is convergent then prove that $$\sum \frac{u_n}{n}$$ is convergent
1983
Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.   For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent.  Not to sure if this is correct, any help would be much appreciated, many thanks.
9374
Suppose that the series $\sum_{n=1}^{\infty}a_{n}$ converges while $\sum_{n=1}^{\infty}a_{n}^2$ diverges. Prove that $\sum_{n=1}^{\infty}a_{n}$ converges conditionally.
8150
How can I prove that this iterative sequence converges to $2$? Can I use the convergence definition?  $$a_{n+1} = \frac{a_n}{2} + \frac{2}{a_n},\qquad  a_0 = 4  $$  Thanks for the help.
31778
Problem:      Suppose that for every $n\in\mathbb{N}$, $a_n\in\mathbb{R}$ and $a_n\ge 0$. Given that   $$\sum_0^\infty a_n$$   converges, show that    $$\sum_1^\infty \frac{\sqrt{a_n}}{n} $$   converges.   Source: Rudin, Principles of Mathematical Analysis, Chapter 3, Exercise 7.
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$$\lim_{n \to\infty} \frac{n!}{n^n}$$  I have a question: is it valid to use Stirling's Formula to prove convergence of the sequence?
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 $0 \leq a_{n} &lt;1$.
41697
Prove that there exists a constant $c&gt;0$ such that for all $x \in [1,\infty)$, $$\sum_{n&gt;x} \frac{1}{n^{2}} \leq \frac{c}{x}$$
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Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
111930
While checking the convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x&gt;0$, we used ratio test to say that for $0&lt; x &lt; \frac1e$ $\sum u_n$ is convergent and for $\frac1e &lt; x&lt;\infty$ $\sum u_n$ is divergent.  We use Logarithimic Test for the case $x = \frac1e$, where we came across computing the limit $$\lim_{n \to \infty} n+n^2 \log \frac{n}{n+1}$$ but I am stuck in finding the limit.
105246
Show that if $a_n$ is descending and $\sum a_n$ is convergent then $na_n \to 0$.  Does $\sum a_n$ have to be convergent?  I see that, but I can't prove it. I think that it need not to be convergent, but i also can't prove it.
111894
This is a homework problem for an undergrad topology course.  Let $l^{2}$ be the set of all real-valued sequences $(c_{n})$ where $\sum_{n=1}^{\infty}c_{n}^{2}$ converges. Let $(a_{n}),(b_{n})\in l^{2}$.  Claim: $\sum_{n=1}^{\infty}|a_{n}b_{n}|$ converges.  I've done a couple of examples where $a_{n}$ and $b_{n}$ are general harmonic series. I've also written out the definition of convergence for $a_{n}$ and $b_{n} $. My problem is that thus far I do not understand on any level why my claim is true.
44577
Show that the series   $$\sum_{n=2}^{\infty} \frac{1}{n\, (\log n)^s}$$ converges for $s &gt;1$ and diverges for $s=1$.
88487
Let we have the following sequence $(a_n)$ such that $$a_n=\sum_{k=1}^n\frac{1}{n+k}$$ How can I prove that $(a_n)$ is increasing bounded sequence, then prove it is convergent and find its limit?
68839
I want to prove the following:  Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences with $a_n&gt;0$ and $b_n&gt;0$ for all $n \geq N$, and let $$c_n = b_n - \frac{a_{n+1}b_{n+1}}{a_n}$$  Then   If there exists $r&gt;0$ such that $0&lt;r\leq c_n \text{ ; } \forall n\geq N$ then $\displaystyle\sum a_n$ converges. If $c_n\leq0$ for $n\geq N$ and if $\displaystyle\sum \dfrac{1}{b_n}$ diverges, then so does $\displaystyle\sum a_n$.   So far I know how to use and prove Cauchy's, D'Alambert's and the integral criterion, so I'd like a hint on either using those or a new idea. The book suggests that for $1$, I show that  $$\sum\limits_{k = N}^n {{a_k} \leq  \frac{{{a_N}{b_N}}}{r}} $$  and for $2$, to prove that $\displaystyle\sum a_n$ dominates $\displaystyle\sum \dfrac{1}{b_n}$  I'd like to prove this with previous theory on series convergence (Cauchy's and/or D'Alambert's preferrably, comparison test) since it is what precedes the problem, and would appreaciate great HINTS rather than answers.   I can't seem to understand the inequalities. I mean, Cauchy's and D'Alambert's criteria reveal that the proof relies on the fact that a geometric series of with ratio $|r| &lt; 1$ converges, but I can't understand the motivation in this proof.    Summing up the work.  For 1:  $$\eqalign{   &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{b_{N + 1}}  \cr    &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{b_{N + 2}}  \cr    &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \frac{{{a_{N + 3}}}}{{{a_N}}}{b_{N + 3}} \cr} $$  Induction over $n$ we get  $${b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}{c_{N + n}} + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}$$  So  $$\eqalign{   &amp; {b_N} \geqslant r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}  \cr    &amp; {b_N} &gt; r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}} \right)  \cr    &amp; \frac{{{a_N}{b_N}}}{r} &gt; {a_N} + {a_{N + 1}} + {a_{N + 2}} +  \cdots  + {a_{N + n}} = \sum\limits_{k = N}^n {{a_k}}  \cr} $$  As desired.
55354
Ch.3 #7. Prove that the convergence of $\sum a_n$ implies the convergence of $$\sum \frac{\sqrt{a_n}}{n},$$ if $a_n \geq 0$.      My attempt. If $\sum a_n$ is convergent, then by the root test, $\lim_{n \rightarrow \infty} sup \sqrt[n]{|a_n|} &lt; 1$. This means $|a_n|^{1/n} &lt; 1$. Because $a_n \geq 0$, this means $a_n ^{1/n} &lt; 1$. Now apply the root test to $\sum \frac{\sqrt{a_n}}{n}$ by considering $$\lim_{n \to \infty} sup \sqrt[n]{|\frac{\sqrt{a_n}}{n}|}.$$ $$\sqrt[n]{|\frac{\sqrt{a_n}}{n}|} = \frac{\sqrt{{a_n}^{\frac{1}{n}}}}{\sqrt[n]n}.$$ Because  $a_n ^{1/n} &lt; 1$, certainly the numerator on the left-hand side of the equation above is $&lt;1$. And because $n$ tends to $\infty$, and $n &gt; 1 \rightarrow \sqrt[n]n &gt; 1,$ the denominator is $&gt;1$. Therefore the limit is $&lt;1$ and the root test implies that $\sum \frac{\sqrt{a_n}}{n}$ converges.     If there are any errors in my proof, I think they come from 1) my assumption of the converse of the root test theorem and 2) my misunderstanding of $\lim_{n\to\infty} sup$. Thanks in advance for your help.
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?
22882
How to prove that  $$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$