Let $ a_n $ be a non-negative sequence, such that $ \sum_{n=1}^{\infty}\sqrt{a_{n}} $ converge. Is it true that  $ \sum_{n=1}^{\infty}a_{n} $ converge? I think that it is. But I want to make sure because it appeared in my final exam. Here is my reasoning: Since $ \sum_{n=1}^{\infty}\sqrt{a_{n}} $ converges, $ \sqrt{a_{n}}\underset{n\to\infty}{\to}0 $ and thus $ a_{n}\underset{n\to\infty}{\to}0 $. So there exists some $ n_0 $ such that, for all $ n&gt;n_0$, it follows that $ 0\leq a_{n}&lt;\frac{1}{2} $. Thus, for each $ n&gt;n_0 $ we have $$ a_{n}&lt;\sqrt{a_{n}} .$$ So, from the comparison test, we get the convergence of $$ \sum_{n=1}^{\infty}a_{n}. $$ Do you agree?
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?
15680
Does $\sum_{n=1}^{\infty} \sqrt[n]{3} - 1$ converge or diverge?   Hints please  I have that $0 \le \sqrt[n]{3}  - 1 \le 1$  Does this help?
116649
How can I show that for any sequence $(a_n)_{n\in\mathbb N}$ and $p&gt;\frac12$, we have that $$\sum_n \frac{\sqrt{a_n}}{n^p}$$ converges if $$\sum_n a_n$$ converges?
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.
77115
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$. Does $\sum_{n=1}^{\infty}a_n$ converge or diverge when compared to $\sum_{n=1}^{\infty}b_n$ if $p &lt; 1$? My answer to this is that it diverges, since $b_n$ diverges if $p&lt;1$, and by a comparison test $a_n$ does as well. Is this correct? Thank you.
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.
112089
Let $(a_n)$ be a sequence such that $(a_n+a_{n+1})$ is convergent. Prove that $(a_n/n)$ converges to $0$.   $(a_n+a_{n+1})$ convergent to $L$ means that for all $\epsilon$, there exists $N$ such that for all $n\geq N$, $|L-a_n-a_{n+1}|&lt;\epsilon$, or in other words $L-\epsilon&lt;a_n+a_{n+1}&lt;L+\epsilon$. How to proceed from here?
39856
I've got a question about the convergence of a series during studying analysis.  If I know that a series of positive real numbers $$\sum_{n=1}^\infty a_n$$ converge, why does $$\sum_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}$$ also converge for $p&gt;1$? Although I know about many convergence tests, I don't know how to apply those tests for this case. Since this problem is the form of "series A converge → series B converge", I've been thinking that it must be verified by using some "comparison" tests. Is this thinking correct? All advice is welcome^_^ Thanks.
104028
Assume that $a_n &gt;0$, $n\in\mathbb{N}$, and that $\sum\limits_{n=1}^\infty a_n $ converges. Show that for $\alpha&gt;1/2$ the series $$\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}$$ converges as well  I think we can use Abel's convergence test but I'm not sure if that is the way to go
66694
I'm currently tackling the following infinite series in order to determine if it converges or not:  $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n}$$  What I've done is rearranged this to make use of the comparison test, shown below:  $$= \frac{\sqrt{n+1}-\sqrt{n}}{n} * (\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$  $$= \frac{(n+1)-(n)}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n^{3/2}+\sqrt{n+1}} &lt; \frac{1}{\sqrt{n+1}} &lt; \frac{1}{\sqrt{n}} $$  However, $\frac{1}{n}$ is smaller than $\frac{1}{\sqrt{n}}$, therefore:  $$  \frac{1}{n}   &lt; \frac{1}{\sqrt{n}}$$   This implies divergence by the comparison test due to the fact $\frac{1}{n}$ diverges, and if a smaller series diverges any larger series must as well.  Is this right? If I've made any mistakes in any part, please tell me!
119969
Suppose $\sum_{n=2}^{\infty} a_n$ is a convergent series. Is the following series convergent as well? $$\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)$$ The part with $n^{a_n}-1 $ looks quite similar to $\sqrt[n]{n}-1$ which behaves like $\ln{n}$, however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
45732
Does this series converge or diverge?  $$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$  I tried the comparsion test with $\sum_{n=1}^\infty$ $1 \over n$ but this does not help.  I also tried to simplify the series to $\sum_{n=1}^\infty \frac{1}{n (\sqrt{n+1}+\sqrt{n}{}{}{})}$ but this become harder.
69177
Do the following series converge if $a_n&gt;0$ and $\sum_{n=1}^{\infty}a_n$ diverges ?  a.) $\sum_{n=1}^{\infty}{a_n \over 1+ a_n}$  b.)$\sum_{n=1}^{\infty}{a_n\over 1+ a_n ^2}$  c.)$\sum_{n=1}^{\infty}{a_n\over 1+ na_n }$  d.)$\sum_{n=1}^{\infty}{a_n\over 1+ n^2a_n }$  What I thought was using Alembert criteria , but it just not seem to work. I also now if $a_n$ converges so does $a_n^2$. Ive tried everything to no avail, can anyone please help ?
727
Setting $a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\ $ I get that $$a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.$$ Clearly all numbers in the sequence are positive and we see that $a_n&lt;a_{n+1} \ \forall \ n$ which implies that the sequence is strictly increasing. To show that it's convergent, I need to show that it's bounded above.  We can use the help-function $f(x)=\sqrt{2+x}$ such that $a_{n+1}=f(a_n).$ But since $$f'(x)=\frac{1}{2\sqrt{2+x}}=0\Leftrightarrow\text{no real solutions,}$$  $f(x)$ never flattens out or decreases, so it can't be convergent?
40706
Let $a_n$ be a sequence. $\lim_{n\to\infty} a_n =0 $ then the series: $\sum_{n=1}^\infty a_n, \ \sum_{n=1}^\infty (a_n+a_{n+1}) $ converge and diverge together.  It doesn't seem right, there are no convergence tests that fit this question yet every counterexample I try does not seem to work. Is this how I should see the sum? $\sum_{n=1}^\infty (a_n+a_{n+1})=2\sum_{n=1}^\infty a_n +a_{n+1}$ If so then there's no real difference between the two series since you multiply it by a constant and add just one more value.
376
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.
32839
Let's suppose that $(a_n)$ is a sequence so that $\sum a_n^2 n^2$ converges, so I have to prove that $\sum |a_n|$ converges.    By the Cauchy criterion we have that, since $\sum a_n^2 n^2 $ converges if $\varepsilon &gt; 0$ then there is a $k\in\mathbb N$ so for all $m,n$ naturals that satisfy $m &gt; n \geq k$: $$ |n^2 a_n^2 + ... + m^2 a_m^2| &lt; \varepsilon $$ but I can't get $| |a_n| + ... + |a_m|  | &lt; \varepsilon$ to show that $\sum |a_n|$ converges. Could you give me a hint? Is there any way to prove this without resorting Cauchy criterion. Thanks!