Let $A_n =\sum \limits_{j=1}^n\frac{\sin j}{j}$. How can I prove that $A_n$ converges as $n$ approaches infinity? I know that if we think of it as an infinite series, it can be proved by applying appropriate convergence tests(for example, Abel's convergence test(or Dirichlet test). But I want a proof not using convergence tests of an infinite series, and think of this as a 'sequence' $A_n$ instead of a 'series'. I read  and other answers related to this question, but most of them used convergence tests and I don't know about Riemann-Lebesgue lemma or Fourier series. Actually I am taking an elementary analysis course, and this is an exercise on the textbook's 'sequence' part, not 'series'.
122827
Question: Find the sum of the series: $$\lim_{n \to \infty}\frac{\sin1}{1}+\frac{\sin2}{2}+\frac{\sin3}{3}+...+\frac{\sin n}{n}$$  I have no clue how to find this. Obviously I can see the sum will be convergent as the denominator gets increasingly bigger while the numerator is bound between $1$ and $-1$.
64925
Let $\{a_n\}$ be a sequence of real number, let $\{n_k\}$ be an increasing sequence of positive integer and let $A_n=a_{n_k}+a_{n_{k+1}}+...+a_{n_{k+1}-1}$ for $k \in N$. Prove that if $\sum_{n=1}^\infty a_n$ converges then $\sum_{k=1}^\infty A_k$ converges. Is the converse true?  I know that $\sum_{n=1}^\infty a_n$ converges implies that $lim _{n_&gt;\infty} a_n =0$, can I conclude that $A_n$ ->$0$ as well, even if I can conclude that, it still doesn't prove that $\sum_{k=1}^\infty A_k$ converges, am I correct?
115884
I ask for some help with this question:  Prove or provide counter example:  If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.  I tries this way:  If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.  There are 3 possible cases:  1) If $a_n &gt;0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}&lt;na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  2) If $a_n &gt;0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}&gt;a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.  Thanks.
5047
I'm not completely new to convergent series and sequences, but proving the above gave me some troubles. Would appreciate some help :) Show that: $$\lim_{n\to\infty} \frac{\sum\limits^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$$
23618
If $a_n \ge 0 $ for all n, prove that $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty {a_n\over 1+a_n}$ converges.   Here is my attempt!  => Suppose that $\epsilon \ge 0$ is given and $a_n$ converges, then  for all $N\le n \le m$$$\sum_{k=n+1}^m a_k \lt \epsilon.$$ let $b_n={a_n\over 1+a_n}$ and Notice that $b_n \lt a_n$ and   Then, applying the comparison test  $$|\sum_{k=n+1}^m b_k| \le \sum_{k=n+1}^m |b_k |\le\sum_{k=n+1}^m a_k\lt\epsilon$$Hence,${a_n\over 1+a_n}$ converges.  let me know if it is worng and I don't know how to start for the other direction. Any help is appreciated!
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?
23840
I just completed the following proof.  Is it valid?   Let $\sum_{k=1}^{\infty} a_k$ be an arbitrary convergent series that also converges absolutely.  Then $\sum_{k=1}^{\infty} a_k \in \mathbb{C}$ and $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$. Let $\{b_k\}$ denote a rearrangement of the $\{a_k\}$.  That is, we have that $\{a_k\} = \{b_k\}$ yet $a_i = b_i$ needn't hold. We will first show that $\sum_{k=1}^{\infty} b_k$ converges absolutely (immediately implying it converges non-absolutely).  We will use the Cauchy Criterion on the partial sums of $\sum_{k=1}^{\infty} |b_k|$ to demonstrate this. First let $\epsilon &gt; 0$. Consider that since $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$, we have that $\exists N_1 \in \mathbb{N}$ s.t. for all $N_1 &lt; m &lt; n$ we have   $$ \sum_{k=m}^n |a_k| \le \epsilon. $$ Now let $0 &lt; N_2$ be large enough s.t $\{a_1, \ldots , a_{N_1}\} \subseteq \{b_1, \ldots , b_{N_2}\}$.  Let $M &gt; max\{N_1, N_2\}$.  Let $M &lt; m' &lt; n'$.  Then since $\{b_k\} - \{b_1, \ldots , b_{m'}\} \subseteq \{a_{N_1}, a_{N_{1}+1},  a_{N_{1}+2} \ldots\}$, we will have that   $$ \sum_{k=m}^n |a_k| \le \epsilon \implies \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon. $$ Then we have that   $$ \left| \sum_{k={m'}}^{n'} |b_k| \right| = \sum_{k={m'}}^{n'} |b_k| &lt; \epsilon $$  so that since $\epsilon$ was arbitrary, we have obtained that   $$ \left| \sum_{k=1}^n |b_k| - \sum_{k=1}^m |b_k|  \right| = \left| \sum_{k=m}^n |b_k|  \right| \rightarrow 0 $$  so that via the Cauchy Criterion for Convergence  we have that $\sum_{k=1}^{\infty} |b_k|$ converges to some  value in $\mathbb{R}$ as desired. Then it follows that $\sum_{k=1}^{\infty} b_k$ is absolutely convergent (and hence also just convergent).   $\square$
83427
I have been trying to solve this problem for awhile, and I can't seem to solve it.Below is the question. I suspect that the series does converge, but I am unable to prove this. Thanks for any help in advance.  Let $(a_n)_{n=1}^\infty$ be a convergent sequence of real numbers. Can it be concluded that $\sum\limits_{n=1}^{\infty}|a_{n+1} - a_{n}|$ converges? Give a proof or a counterexample.
113507
Let $S_n:= \displaystyle\sum_{k=1}^n \displaystyle\sum_{j=1}^{k-1}\dfrac{(-1)^k}{[j(k-j)]^p} $. Determine whether $\lim\limits_{n\to\infty} S_n =\displaystyle\sum_{k=1}^\infty \displaystyle\sum_{j=1}^{k-1}\dfrac{(-1)^k}{[j(k-j)]^p}$ converges or diverges for $\frac{1}2 &lt; p \leq 1.$   I claim that $S_n$ converges as $n\to \infty.$ Let $c_k:= \displaystyle\sum_{j=1}^{k-1}\dfrac{1}{[j(k-j)]^p}&gt;0\,\forall k\in\mathbb{N},$ with $c_1 := 0.$ We want to show that $c_k$ is eventually nonincreasing (ie. $\exists N\in\mathbb{N}$ such that $\forall k\geq N, c_{k+1}\geq c_k$) and that $\lim\limits_{k\to\infty}c_k=0,$ which will show by Leibnitz's alternating series test that the series converges.  However, this seems very difficult to do.      I want to avoid using "well-known" theorems, and instead use more fundamental theorems like Leibnitz's alternating series test, the Cauchy product formula for absolutely convergent series, etc.
46841
Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.   I showed it using Abel's theorem and limit comparison test. Any other simpler method?
45932
Let$a_n,b_n\in\mathbb R$ and $(a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}$. The series $\sum_{n=1}^\infty\frac{a_n}{b_n} $ and $\sum_{n=1}^\infty(\frac{a_n}{b_n})^2 $ are convergent.  How to prove that$$\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} $$ is convergent.  Thanks in advance
124270
Suppose that the series $\sum \limits_{n=1}^{\infty}\dfrac{a_n}{n^{\sigma}} \quad(\sigma&gt;0)$ converges. Prove that $\lim \limits_{n\to \infty}\dfrac{a_1+a_2+\dots+a_n}{n^{\sigma}}=0$.  I applied the Abel transformation, but unsuccessfully.
53321
I'm trying  to find out now whether the series $\sum_{n=2}^{\infty } a_{n}$ converges or not when $$a_n = \frac{n^{\log n}}{(\log n)^{n}}$$  Again, I tried d'Alembert $\frac{a_{n+1}}{a_{n}}$,   Cauchy condensation test $\sum \limits_{n=2}^{\infty } 2^{n}a_{2^n}$, and they both didn't work for me.  I can't use Stirling, nor the integral test.  Edit: I'm searching for a solution which uses sequences theorems and doesn't involve functions.    Thank you
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
64951
Let $(a_n)_{n\in\mathbb{N}}$ be a sequence such that lim inf $|a_n|=0$. Prove that there is a subsequence $(a_{n_{k}})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^\infty a_{n_k}$ converges.  I'm thought about showing that you can make a subsequence that is smaller then $1/n^2$. Is this the right strategy ? Any hints ? Should I use the Cauchy criterion ?
104844
Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.  Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon &gt;0$, for sufficiently large $n$, $|y_n -L| &lt; \varepsilon$ .  But $|y_n - L| &lt; 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$  But then how to proceed?
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!