Let $\{a_n\}$ be a sequence of positive real numbers such that $a_1=1,a_{n+1}^2-2a_na_{n+1}-a_n=0\,\,\forall n\ge 1$. Then show that $$1&lt;\sum_{n=1}^\infty {a_n\over 3^n}&lt;2$$. I got this problem from a question paper. Though it didn't asked for exact sum of the series , if you also manage to find that then please post it.
122212
Let $\{a_n\}$ be a sequence of positive real numbers such that   $a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.   Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...  (A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.   Solution attempt:   Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n&gt;0$, the other root is rejected).   We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}&gt;1$, then $\lim a_n \to \infty$.   Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}&gt;0$. (Again, the other root is rejected due to the same reason).   Hence, $(a_n)$ increases monotonically.   Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$  Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}&lt; 1.3$).  So, option $(A)$ is the correct answer.   Is the procedure correct?   I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.   Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
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?
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?
87463
Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.  Prove that $(a_n)$ is convergent and find the limit.    I have no idea how to prove convergence. I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.  If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.
73031
Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.  I want to prove the following inequality using mathematical induction:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?
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?
41122
I'm having trouble with proving the following for my math study:  Let $x$ be a real number with $|x| &lt; 1$, and $q$ be a real number. Show that the series $\sum_{n = 1}^\infty n^qx^n$ is absolutely convergent, and that $\lim_{n \rightarrow \infty}$ $n^qx^n = 0$  I tried this:  $\sum_{n = 1}^\infty n^qx^n$ = $x + 2^q + 3^qx^3 + ...$ $= x(1 + 2^qx + 3^qx^2 + ...)$  So far, I don't know how I have to do this, I tried to apply the root test on this, but for that I would need to compute the lim sup of the series, so I thought that was not correct. Could you explain me how to prove this?  Thanks in advance!
117044
Show that the series converges, but not absolutely. $\sum_{n=1}^{\infty}( $exp$(\frac{(-1)^n}{n})-1)$.   My Try:  Let $a_n=$exp$(\frac{(-1)^n}{n})-1$. I was going to use alternating series test because the sequence $\{a_n\}$ is alternating. But $\{|a_n|\}$ is not decreasing to $0$. So, I am stuck now. Can anybody please give me a hint?
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
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?
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.
58572
Let $\sum_{n=1}^{\infty}{a_n}$ be a series such that for each $n$, $a_n&gt;0$ and $(a_n)$ is decreasing. Suppose that $\sum_{n=1}^{\infty}{a_{2n}}$ converges. Prove that $\sum_{n=1}^{\infty}{a_n}$ also converges. I try to prove by using definition but I got nowhere . Can anyone guide me ?
83494
If $\sum_{n=2}^{\infty}a_n$ converge, then also converge this series? $$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)$$  Please verify my answer below  Counterexample:  $$a_{n}=\begin{cases} \frac{1}{k^{2}} &amp; n=k!^{k^{2}}\\ \\ 0 &amp; \text{All other cases} \end{cases}$$  When our infinites sum is equal to  $$\sum_{k=1}^{\infty}\frac{1}{k}\cdot\frac{1}{2\, \ln\, k}\cdot(k!-1)$$
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).   $\square$
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?
17821
Given that $\{\phi_n\}$ is a sequence of non-negative numbers whose sum is $1$ and $\{\psi_n\}$ is a sequence of positive numbers, how can I show that  $$ \prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n~? $$    Thanks.    PS: I'm not too sure about the title. Perhaps, someone could give it a better title?
97356
I'm having some difficulty solving this problem :     Given the sequence $(a_n)$ with $a_n \geq a_{n+1} \geq 0$ show that if $$\sum \sqrt{a_n a_{n+1}} &lt; \infty \implies \sum a_n &lt; \infty.$$    I first thought of using the Arithmetic Mean-Geometric Mean Inequality:   $$\forall{a,b} &gt; 0: \frac{a+b}{2} \geq \sqrt{ab}.$$  Therefore, since $a_n \geq a_{n+1} \geq 0$ we have: $$\frac{a_n+a_{n+1}}{2} \geq \sqrt{a_n a_{n+1}} \;\; \forall{n} \in \mathbb{N}.$$  but I can't use the Direct Comparaison Test here since it will only work to show the other side of the implication $\sum a_n &lt; \infty \implies \sum \sqrt{a_n a_{n+1}} &lt; \infty$.