Let $\mathbb {S} =\left \{ 1,2,3,...,9,11,12,...,19,21,...99,111,112,113... \right \} $ i.e, the positive integers set which contain zero digit is omitted.  Now show that $ \sum_{n\in \mathbb {S}}  \frac{1}{n} $ is convergent .  I really don't have no idea about how to prove this
122278
I'm working on the following contest math problem:     Define an evil number to be any positive integer that contains the   digit $9$. Show that      $$ \sum_{x} \frac{1}{x} &lt; 80 $$      where the sum is over all non-evil positive integers $x$.   I'm very confused on where to begin. Initially, I tried to consider this sum as part of the sum of $1/x$ over all positive integers, namely by noting that  $$ \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{x} \frac{1}{x} + \sum_{y} \frac{1}{y} $$  where the first summation on the right side is the same as in the problem statement and the second is the sum over all evil numbers. However, I can't seem to find a way to use this since the sum of the left diverges.  Could anyone lend a helping hand?
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?
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.
26008
Suppose $\sum{a_n}$ converges. How do I prove that $\sum{\frac{\sqrt{a_n}}{n}}$ converges.  So I know that just because  $\sum{a_n}$ converges, do not mean I can say anything about the converges of its square root. So I know that I can prove that if  $\sum{\sqrt{a_n}}$ converges then $\sum{\frac{\sqrt{a_n}}{n}}$ converges by Abel's Test, but I do not know where to start for the case where  $\sum{\frac{\sqrt{a_n}}{n}}$ diverges
48471
For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges.   This is what I have so far..  Let $\epsilon &gt; 0$ be given to us. We must show that there exists an $N \in \mathbb{N}$  such that $n\ge N \implies \left|\frac{1}{2n} - 1\right| &lt; \epsilon$. Choose $N$ to be any positive integer which is larger than $\frac{1}{\epsilon}$. (So $N&gt;\frac{1}{\epsilon}$.) Then $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$
109643
Does $\sum\frac{\sin n}{n}$ converge?  I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent.
9802
Show that  $$\lim \frac{1}{n} \sum_{i=1}^n \frac{1}{i} =0 $$  I've proved that this sequence converges (it is bounded and decreasing). NOW, I need to find a sequence that is bigger than this one and goes to zero. Maybe something using geometric serie of 1/2  Thanks in advance!
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.
70232
I'm asked to find $\displaystyle\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$.  It seems that the sum converges.  I managed to prove that $\forall n,1&lt;\sum_{k=1}^n \left(\frac{k}{n}\right)^n&lt;e$  But I don't know how to prove convergence, nor find the value of the limit.
8475
Can someone give an explanation using the definition of convergence in partial sum to show how the above infinite sum converges to 1?  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
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 $(0, 1)$.
8737
Suppose $a_n \ge 0$ for all natural numbers $\Bbb{N}$ and $\sum{a_n}$ converges. Is it necessarily true that $\sum{(a_n)^2}$ converges?  I think yes.  Clearly as $a_n \ge 0$ it follows $(a_1+a_2+\dots+a_n)^2 \ge a_1^2+a_2^2+\dots+a_n^2 \ge 0$ and so $(\sum_{n=1}^{\infty}{a_n})^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$.  As we are told $\sum{a_n}$ converges to say $L$ where $L$ is a real number it follows $L^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$ and so therefore $\sum{(a_n)^2}$ converges.  Is this correct?
55167
let $m$ is give a positive integers,   Determine for which values of $a$,the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}$$ converges     where $[x]$ is the largest integer not greater than $x$.  and  I have see this not hard problem :show that $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges, solution: note     $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}=\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}\right)$$   and we have   let $$a_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}$$   $$0&lt;a_{n}&lt;\dfrac{2n+1}{n^2}&lt;\dfrac{3}{n}\longrightarrow 0$$   and   $$a_{n}-a_{n+1}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}+\dfrac{1}{n^2+1}-\dfrac{1}{(n+1)^2+1}+\cdots+\dfrac{1}{n^2+2n}-\dfrac{1}{(n+1)^2+2n}-\left(\dfrac{1}{(n+1)^2+2n+1}+\dfrac{1}{(n+1)^2+2(n+1)}\right)$$   so   $$a_{n}-a_{n+1}=(2n+1)\left(\dfrac{1}{n^2(n+1)^2}+\cdots+\dfrac{1}{(n^2+2n)(n^2+4n+1)}\right)-\dfrac{1}{n^2+4n+2}-\dfrac{1}{n^2+4n+3}$$   so   $$a_{n}-a_{n+1}&gt;\dfrac{(2n+1)^2}{(n^2+2n)(n^2+4n+1)}-\dfrac{2}{n^2+4n+1}=\dfrac{2n^2+1}{(n^2+2n)(n^2+4n+1)}&gt;0$$   so   $$\sum_{n=1}^{\infty}(-1)^na_{n}=\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges   But for my problem,I can't   My try:  $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}=\sum_{n=1}^{\infty}(-1)^n\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$ let $$b_{n}=\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$ then how prove  $$b_{n}\longrightarrow 0$$and $$b_{n}&gt;b_{n+1}$$
62417
suppose that $\sum_{n=1}^{\infty}a_n$ is a convergent series of positive terms. Prove that $\sum_{n=1}^{\infty} \sqrt{a_n \cdot a_{n+1}}$ is also convergent. Demonstrate that the converse is false.
64190
Given that the series $(a_n)$ converges absolutely. Show that the series $(\frac{a_n}{1 + a_n})$ converges absolutely.    I am not really sure where to start. Any help would be great.
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.