Let $a_n$ be a sequence of real numbers such that $\sum_{n=1}^{\infty} a_nb_n &lt; \infty$ for any sequence $b_n$ satisfying $\sum_{n=1}^{\infty}b_n^2 &lt; \infty$. Prove that $\sum_{n=1}^{\infty}a_n^2 &lt; \infty$.  This question is in the chapter on Hilbert spaces. What sort of technique is required here? Seems like a baby analysis question.
11822
I'm trying to prove the following:     If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n&lt;\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2&lt;\infty$, then $\sum_{n=1}^\infty a_n^2 &lt;\infty$.   The context here is functional analysis homework, in the subject of Hilbert spaces.  Here's what I've thought:  Let $f=(a_n)&gt;0$. Then the problem reads: if $\int f\overline{g}&lt;\infty$ for all $g&gt;0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.  I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle &lt;\infty$ for all $g&gt;0,g\in \ell^2$ then $f\in \ell^2$.  I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P  From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$  but it's not useful here.
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.
56679
a) Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.  I wanted to use the comparison test to show it's true, but I think I got something mixed up. Here's what I have so far.  $b_n$ is bounded $\Rightarrow \exists c &gt; 0: |b_k| &lt; c, \forall k \in \mathbb N$  $\sum a_nb_n &lt; \sum a_nc &lt; c\sum a_n$. I'm very tempted to use the comparison test here and say, as $\sum a_n$ converges absolutetly, then so does $c\sum a_n$, but for that I needed the inverted relation, right? I would need $\sum a_n &gt; c\sum a_n$, which is not true. However isn't it obvious that something absolutely convergent multiplied by a constant is also absolutetly convergent? Is there a mathematical way to write this? Thanks a lot in advance.  b) Refute with a counter example: if $\sum a_n$ converges and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges.  Is this even possible? I mean, it has to be, but it doesn't make sense to me. If something converges, it means it's bounded, right? And I thought, well, since bounded + bounded = bounded, then bounded*bounded would also get me something bounded again.  Anyway, my idea would be to use for $a_n$ an alternating series that ist only convergent, for example $(-1)^n \frac 1 {n^2}$. But something tells me I'm trying to prove $a_n b_n$ is not absolutely convergent.  Thanks a lot in advance guys!
100666
Let $(a_n)_{n=1}^{\infty}$ be a sequence, let $\; 0 &lt; q &lt; 1$ such that $\;\forall n \in \mathbb{N}$  $\;n&gt;2$    $ \left | a_{n+1} -a_n \right | &lt; q\left | a_{n} -a_{n-1}\right |$  prove that $(a_n)_{n=1}^{\infty}$ converges.  Well, I thought about proving it this way:  $ \left | a_{n+1} -a_n \right | &lt; q*\left | a_{n} -a_{n-1}\right |&lt;\left | a_{n} -a_{n-1}\right | $    therefore $ \left | a_{n+1} -a_n \right | &lt;\left | a_{n} -a_{n-1}\right | \;$  for each $n&gt;2$ and that obviously shows us that the sequence converges because the gaps between each $2$ indexes reduces as $n$ grows.  I have a feeling however that this is not a formal way of proving this, it seems too easy.
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?
53291
Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct?
65491
Let $(x_n)_{n=1}^{\infty}$ be a sequence such that $|x_{n+1} - x_n| &lt; r^n$ for all $n \geq 1$, for some $0 &lt; r &lt; 1$. Prove that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.  I understand that a Cauchy sequence means that for all $\varepsilon &gt; 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. But this one is really giving me a headache.  I tried doing something like: let $ m &gt; n$. Therefore $x_n - x_m$ =  $(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) $ and then somehow using the triangle inequality to compute some sum such that $x_n - x_m$  &lt; sum which would be epsilon?  any help is appreciated, thank you.
61090
Let $(a_n)_{n=0}^{\infty}$  be a sequence of real numbers satisfying $\sum_{1}^{\infty} |a_n - a_{n-1}| &lt; \infty$   Which of the following conclusions must then be true?  The series $\sum a_nx^n$ converges   Nowhere on $ \mathbb{R}$ Everywhere on $  \mathbb{R}$ On some interval containing $(-1,1)$ Only on $(-1,1)$   What I have tried is Make an inequality with $|a -b| \leq |a| +|b|$ .. but i cant proceed further .. Any sort of help will be deeply appreciable .
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.
30158
Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that      If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero.    A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.  Could anyone please help me to prove it?
44043
We let a>0 and b>0 and  we have $\sum_{n=0}^{\infty}b^n-a^n$. We have to find all values a,b>0 where $\sum_{n=0}^{\infty}b^n-a^n$ is convergent. I think I can use that $$\sum_{n=0}^{\infty} b_n &lt; \infty , \sum_{n=0}^{\infty} a_n &lt; \infty \implies \sum_{n=0}^{\infty} (b_n-a_n) &lt; \infty. $$. And I think $\sum_{n=0}^{\infty}b^n$ and $\sum_{n=0}^{\infty}a^n$ is convergent if a&lt;1 and b&lt;1? It will make sense. But how can I show it?
88825
$\sum_{n=1}^\infty n*(n+1)*x^n$  Hello everyone,  I need help in solving the question above.  I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue. Thanks.
61698
This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.
71569
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:  Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.  I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.
19474
Let $(x_n) _{n\ge 1}$ be a sequence of real numbers such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$. Prove that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=0$. I used the definition of the limit to conclude that $\exists N\in \mathbb{N} $ such that $|\frac{x_1^2+x_2^2+...+x_n^2}{n}|&lt;\frac{1}{n^2}$, $\forall n\ge N$. Hence, we get that $|x_1^2+x_2^2+...+x_n^2|&lt;\frac{1}{n}$. Now here comes the part where I am not really sure. I think that this implies that $\sum_{n=1}^{\infty}x_n^2=0$, and as a result $x_n\to 0$,which solves the problem because if we use the Stolz-Cesaro lemma we get that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=\lim\limits_{n\to \infty} x_n=0$.
28033
Proof that $\lim{a_n}=L$ when $n$ goes to infinity, then $\{a_n\}$ its a Cauchy Sequence  I start with this hypothesis  $\displaystyle\lim_{n \to\infty}{a_n}=L \Leftrightarrow{\forall{\epsilon}&gt;0}$  $\exists{N}\in{\mathbb{N}}$ such that  $\forall{n}&gt;N \left |{a_n-L}\right |&lt; \epsilon$  The same its true fot any $m&gt;N,\left |{a_m-L}\right |=\left |{L-a_m}\right |&lt; \epsilon$  I will sum the inequalities but I havent any idea how do it... Help me please!!!
6372
If $$\sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}}$$ Converges( s is real) and $\operatorname{Re}(z)&gt;s$.  Then $$\sum_{n=1}^{\infty} \frac{a_{n}}{n^{z}}$$ also converges.  $a_n$ is complex sequence.