Let $\{a_n\}_{n=1}^{\infty}$ be a sequence with non negative terms so that for every sequence $\{x_n\}_{n=1}^{\infty}$ with $x_n \geq 0$ and $\lim_nx_n=0$, the series $\sum_{n=1}^{\infty}x_na_n$ converges. Show that $ \sum_{n=1}^{\infty}a_n $ also converges.  We can proceed through contradiction by assuming $ \sum_{n=1}^{\infty}a_n = \infty$ and then constructing a sequence $\{x_n\}_{n=1}^{\infty}$ such that  $ \sum_{n=1}^{\infty}x_na_n $ diverges.
69333
A friend of mine gave me the following problem recently:     Let $(a_n)$ be a sequence of real numbers. Suppose that $\sum_{n=1}^\infty a_nb_n$ converges for any sequence $(b_n)$ such that $\lim\limits_{n\to\infty} b_n=0$. Show that $\sum |a_n|&lt;\infty$.   Hint he gave me was to use  (a.k.a. uniform boundedness principle).   I would like to know whether there are different solutions to this, in particular, whether there is an elementary solution. (Using just calculus, without resorting to more advanced results from functional analysis or other areas.)     The argument using Banach-Steinhaus theorem goes as follows:  Let $a=(a_n)$ be a given (fixed) sequence such that $\sum a_nb_n$ converges for each $b=(b_n)\in c_0$.  For every $n$ we can define $f_n\colon c_0\to{\mathbb R}$ as $\sum_{k=1}^n a_nb_n$. We consider $c_0$ with the usual sup-norm, which makes $c_0$ into a Banach space.  Each function $f_n$ is obviously linear.  For any $b\in c_0$ such that $\|b\|=\sup\limits_{n\in\mathbb N} |b_n| \le 1$ we have $$|f_n(b)| \le \sum_{k=0}^n |a_k|.$$ This shows that all functions $f_n$ are bounded linear functionals.   Now if we fix some $b=(b_k)\in c_0$ we have $$\left|\sum_{k=0}^n a_kb_k\right| \le \sum_{k=0}^\infty |a_k|\cdot |b_k|.$$ The sum on the RHS is finite, since it can be expressed as $\sum a_kc_k$, where $c_k=\pm b_k$. (The choice of sign here depends on the signs of $b_k$ and $a_k$.) So we have shown that at each point $b\in c_0$ the values $f_n(b)$ are bounded by the same constant (depending only on $b$).  By Banach-Steinhaus theorem we have that there exists a constant $M$ such that $\|f_n\|&lt;M$ for each $n\in\mathbb N$.  Now it only remains to show that $\|f_n\|=\sum_{k=0}^n |a_k|$. We already know that $\|f_n\|\le\sum_{k=0}^n |a_k|$. The other inequality can be obtained by choosing $b$ in unit ball, such that $b_k$ is either $1$ or $-1$, depending on the sign of $a_k$ for $k=1,2,\dots,n$; and $b_k=0$ for $k&gt;n$.  Now we have that $\sum_{k=0}^n |a_k|&lt;M$ for each $n$, which implies $\sum_{k=0}^n |a_k|\le M$.
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.
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?
8578
Let $a_n \subset \mathbb C$ is a sequence Show that if $\sum_{n=1}^{\infty} a_nx_n \lt \infty$ $\forall x_n \in \ell^1$   then $a_n \in \ell^{\infty}$  What did I do : $x_n \in \ell^1 \Rightarrow \sum_{n=1}^{\infty} |x_n| \lt \infty $ Since $\mathbb C$ is a Banach space $\sum_{n=1}^{\infty} x_n \lt \infty $ and from Leibnitz rule $x_n \to 0$ $\sum_{n=1}^{\infty} a_nx_n \lt \infty \Rightarrow$ $a_nx_n \to 0$ i.e. $\forall \varepsilon \gt 0$$ |a_nx_n| \lt \varepsilon $ $|a_n||x_n|=|a_nx_n| \lt \varepsilon \Rightarrow$ $|a_n| \lt \varepsilon /|x_n| $ and $|x_n|$ is a finite number thus terms of $a_n$ are bounded. So $a_n \in \ell^{\infty}$ I’m not sure about my writtens. It seems wrong to me. Could someone fix me please? Thanks a lot
28920
Prove that if the positve term series $\sum^{\infty}_{n=1}a_n$ is convergent, also $\sum^{\infty}_{n=1}\sqrt{a_na_{n+1}}$ is convergent.  Prove that if the positive term series $\sum^{\infty}_{n=1}a_n$ and $\sum^{\infty}_{n=1}b_n$ are convergent, also $\sum^{\infty}_{n=1}a_nb_n$ is convergent.  I've tried to solve it using comparison test, but no results.
21003
Prove or disprove: if $\sum\limits_{n=0}^\infty a_n$  converges and $\lim\limits_{n→\infty}⁡b_n =0$, then $\sum\limits_{n=0}^\infty a_n  b_n$ converges.  I tried this example $$a_n= b_n = \frac {(-1)^n} {\sqrt {n+1}},$$ this will show that $$\sum_{n=0}^\infty a_n  b_n = \sum_{n=0}^\infty \frac {1} { {n+1}},$$ which is the Harmonic series, diverges.   However, I tried every convergent test but I couldn't show that $$\sum_{n=0}^\infty \frac {(-1)^n} {\sqrt {n+1}}$$ converges.
115884
I ask for some help with this question:  Prove or provide counter example:  If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.  I tries this way:  If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.  There are 3 possible cases:  1) If $a_n &gt;0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}&lt;na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  2) If $a_n &gt;0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}&gt;a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.  Thanks.
83427
I have been trying to solve this problem for awhile, and I can't seem to solve it.Below is the question. I suspect that the series does converge, but I am unable to prove this. Thanks for any help in advance.  Let $(a_n)_{n=1}^\infty$ be a convergent sequence of real numbers. Can it be concluded that $\sum\limits_{n=1}^{\infty}|a_{n+1} - a_{n}|$ converges? Give a proof or a counterexample.
118736
$p&gt;1$ is a integer, Show  a  convergent series $\sum\limits_{n=1}^\infty a_n$, $a_n\in\Bbb R$, such that the series  $$\sum_{n=1}^\infty a_n^p$$  is divergent  p.s. If $p&gt;1$ is not an integer and $a_n\lt0$, it will be difficult to define $a_n^p$(complex analysis?), so we only consider $p\in\Bbb Z$
52022
This problem is form the Romanian GM.  Let $(x_{n})_{n\geq1}$ be a sequence of real numbers defined by $x_{1}=1$ and $$x_{n+1}=x_{n}+ \frac{\sqrt{n}}{x_{n}}$$ for all $n \geq 1$. Study the convergence of the sequence $$(\frac{x_{n}}{n^{\frac{3}{4}}})_{n\geq1}$$  I only managed to show from the given relation that $(x_{n})_{n\geq1}$ is strictly increasing and tends to $+\infty$.
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Given a positive sequence $(a_n)$. Prove that if $\displaystyle \sum_{n=1}^{\infty} a_n$ converges, then so does $\displaystyle \sum_{n=1}^{\infty} a_n^{\frac{n}{n+1}}$.   I tried to use the root test, and I was stuck with the case $\lim \sup \sqrt[n]{a_n}=1$. Maybe we need some trick here.  Thanks so much.
23900
Per the title, if $\sum_{1}^{\infty}(a_n)^3$ diverges, does this imply that $\sum_{1}^{\infty}(a_n)$ diverges?   I'd appreciate hints (!) for dealing with this excercise.  EDIT Per the contrapositive, it is not given that $a_n$ converges absolutely, or that it is nonnegative for all $n$.  Thank you!
376
Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
30158
Let $a_n$ be a sequence of nonnegative numbers. Then I want to show that      If $\,\sum_{n=1}^\infty a_n^2\,/ n\,$ converges, then $\frac 1N\!\sum_{n=1}^{N}a_n\,$ converges to zero.    A hint suggests me to write $a_n=\sqrt{n}\cdot \frac{a_n}{\sqrt n}$ and use the Cauchy-Schwarz inequality. However it does not seem enough. I cannot show that $\frac 1N\!\sum_{n=1}^{N}a_n$ converges to zero.  Could anyone please help me to prove it?
38539
Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon &gt; 0$ do we have $$\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 $(0, 1)$.
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
96153
Let $a_n$ be a positive sequence such that $S_n = \sum\limits_{k=1}^n a_k$ diverges.   I'm trying to prove $\sum\limits_{k=1}^n \frac{a_n}{S_n}$ diverges.   I tried summation by parts, limit comparison and Stolz theorem in many different combinations and am still stuck with nothing.