If $\sum a(n)$ and $\sum b(n)$ are convergent series with positive terms,can I say that $\sum a(n)b(n)$ is also convergent?
123884
Given $a_n, b_n$ such that $\exists N   \forall n&gt;N  a_n, b_n \geq 0$ I want to show that if $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge then $\sum_{n=0}^{\infty}a_n b_n$ also converges.  I tried to show it in the following way, if $\sum_{n=0}^{\infty}b_n$ converges, then $\lim_{n\rightarrow\infty}b_n=0$, so there exists $n_0$ such that $b_{n_0}\geq b_n$ for all $n\gt n_0$, therefore $a_n b_{n_0}\geq a_n b_n$, so that $\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n$, and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning?
68121
Does the following series converge or diverge?  $\sum_{n=1}^{\infty} a_n$ where  $a_n = \frac 1 b_n$, and $(b_n)_n$ is the subsequence of $(n)_n$ whose terms do not have a $9$ in their decimal notation.  I have no idea how the sequence $b_n$ even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what $b_n$ is.
38539
Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} &lt; \infty?$$  For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon &gt; 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$.
118919
If $a_n &gt;0$ for all $n\geq1$, show that the series  $$\sum_{n=1}^{\infty} \frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}$$  converges.  Can someone check if my solution is correct:  $$\sum_{n=1}^{M}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_M)} &lt; 1 $$  Since the partial sum of the series is bounded, the series is convergent.
4075
I have to prove the  by tomorrow and I am really unconfident about what I did:  $$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$  I did the following:  Let $(b_n)$ be a sequence as follow: $b_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $a_{n+1}\leq a_n$ and so $0\leq a_{n+p}\leq a_n$ for all $n,p\in\mathbb N$.  So $\sum\limits_{n=1}^\infty b_n$ converges by the majorizing series $\sum\limits_{n=1}^\infty a_n$. And it's $\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}$ so $\Rightarrow$ is done.  For $\Leftarrow$ consider $c_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m&lt;2^k$.  It's $|a_n|\leq c_n$ and $\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}$ and so $\sum\limits_{n=1}^\infty a_n$ converges by the majorizing series $\sum\limits_{n=0}^\infty c_n$.  Is this in form and content correct?
40393
If $\sum a_n$ with $a_n&gt;0$ is convergent, then is $\sum {a_n}^2$ always convergent? Either prove it or give a counter example.  Im trying in this way, Suppose $a_n \in [0,1] \ \forall\  n.\ $ Then ${a_n}^2\leq a_n\ \forall\  n.$ Therefore by comparison test $\sum {a_n}^2$ converges.  So If $a_n$ has certain restrictions then the result is true.  what about the general case?  How to proceed further? Hints will be greatly appreciated.
34945
I am sorry that my question is not whether $\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}$ convergent or not.  My question is how to compute the sum of series:  $$\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},$$ It is a question in a textbook, and I have worked on it for days, but fail.
82381
Is the following series convergent?  $$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$  I treid the Ratio and Root tests, but both of them failed.
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?
109414
If we are given any sequence of real numbers $\{a_n\}$ diverges, so $a_n \rightarrow + \infty$, how can we find a sequence $\{b_n\}$ such that $\sum |b_n|$ converges but $\sum |a_n||b_n|$ diverges?  I want to use this fact in another problem but don't immediately see how to prove it.
1293
I need help to solve following problem from Rudin's Mathematical analysis book:    Convergence of the series $\sum a_n$  implies the convergence of $\sum \dfrac{\sqrt {a_n}}{n}$, if $a_n&gt;0$  I tried to construct a suitable convergence sequence $b_n$ such that $\sum b_n$ converges and $a_n \leq b_n$ but, I am not able to find such sequence $b_n$ .  Thanks for the help and sugestions.
7729
Let $\displaystyle \sum a_n$ be  convergent series of positive terms and set  $\displaystyle b_n=\sum_{k=n}^{\infty}a_k$ , then prove that $\displaystyle\sum \frac{a_n}{b_n}$ diverges.   I could see that $\{b_n\}$ is monotonically decreasing sequence converging to $0$ and I can write $\displaystyle\sum \frac{a_n}{b_n}=\sum\frac{b_n-b_{n+1}}{b_n}$, how shall I proceed further?
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.
115363
We have two infinite series $\sum a_n $ and $\sum b_n$, such that $a_n &gt; 0,\ b_n &gt; 0$ and $a_{n + 1}/a_n \leq b_{n + 1}/b_n$ for all $n$. Show that if $\sum b_n$ converges, then $\sum a_n$ also converges.   Any hints on how to prove this? I can sort of picture why this must be true, but I'm not sure how to prove this rigorously. I'm not sure using the ratio test would help here..
33086
Let $\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$,$s_k$ denotes the partial sum then could anyone tell me which of the following is/are correct?  $1.$ $s_k=0$ for infinitely many $k$  $2$. $s_k&gt;0$ and $&lt;0$ for infnitely many $k$  3.$s_k&gt;0$ for all $k$  4.$s_k&gt;0$ for all but finitely many $k$  if we take $a_n=(-1)^n{1\over n}$ then $\sum a_n$ is convergent but not absolutely convergent,but I don't know $\sum_{n=1}^{\infty} a_n=0$? so I am puzzled could any one tell me how to proceed?
113494
Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..
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?