I need to prove or disprove the statement. I think the statement is true.  My attempt at a proof:  From the definition of convergence:  $$\forall \epsilon &gt; 0 \quad \exists N \in \mathbb{N} \quad st \quad n,m\geq N \implies \left|\frac{a_{m+1}}{1+a_{m+1}}+\dots+\frac{a_{n}}{1+a_{n}}\right|&lt;\epsilon$$  Since there is a finite amount of denominators and $a_n&gt;0$ we can take the maximum:  $$g = \max\{1+a_{m+1}, \dots, 1+a_n\}$$  $$\frac{1}{g}\left|a_{m+1},\dots,a_n\right|\leq\left|\frac{a_{m+1}}{1+a_{m+1}}+\dots+\frac{a_{n}}{1+a_{n}}\right|&lt;\epsilon$$  and that completes the proof since $\epsilon$ is arbitrary. Does this make sense?
55762
Let $\sum\limits_{n\geq1}a_n$ be a positive series, and $\sum\limits_{n\geq1}a_n=+\infty$, prove that: $$\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.$$
9450
Let $(X,d)$ be a metric space, and say $(x_n)$ is a Cauchy sequence such that it has a convergent subsequence $(x_{n_k})$ that converges to $x$. We show $x_n \to x$. Let $\epsilon &gt; 0$. Take $N &gt;0$ such that for all $n,m &gt; N$, we have   $$d(x_n,x_m) &lt; \frac{\epsilon}{2}.$$  By hypothesis, we can take also $K &gt;0$ such that for all $n_k &gt; K$, we have  $$ d(x_{n_k},x) &lt; \frac{\epsilon}{2}.$$  Put $M = \max \{N,K\}$. Therefore, for all $n,m,n_k &gt; M$, we have  $$ d(x_n,x) \leq d(x_n, x_{n_k}) + d( x_{n_k},x) &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$  Hence, $x_n \to x$ as desired.  Is this a correct approach? Thank you very much in advance.
47290
This is the other direction of  that reads  Let $a_n&gt;0$ (!) for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty a_n&lt;\infty$ implies $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}&lt;\infty$.  I need to prove that &quot;the converse of the last statement is false&quot;. I don't want the answer, but to ask if I can use the following argument to prove the statement. To start, I'm confused with the statement &quot;the converse of the last statement is false&quot;. Is it enough if I prove, if $\sum_n \sqrt{a_n a_{n+1}}$ diverges, then $\sum_n a_n$ must diverge? If this is true, then the proof is easy since I can choose $a_n=1/n$ and prove that $\sum_n a_n$ converges.
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?
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?
80831
I'm reading through a proof in my lecture notes:     Every convergent sequence of a metric space has a unique limit.   Suppose $\lim_{n \to \infty} x_n = l$ and $\lim_{n \to \infty} x_n = m.$ Then for every $\epsilon &gt; 0,$ there exists $N$ such that for every $n \geqslant N,$  $$d(l,m) \leqslant d(l,x_n) + d(x_n,m) &lt; \epsilon + \epsilon = 2 \epsilon. $$  So $(\star) \hspace{1mm} 0 \leqslant \frac{d(l,m)}{2} &lt; \epsilon \Longrightarrow \hspace{1mm} (\star^\prime) \hspace{1mm} \frac{d(l,m)}{2} =0 \Longrightarrow l = m.$  I can follow the proof up until the jump from $(\star)$ to $(\star^\prime)$. Why does $\frac{d(l,m)}{2}=0$?
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
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 $g=1$.
115740
I've been working on the following problem:      Show that if $f\in C[a, b]$ , $f\ge 0$ on $[a, b]$, then $\left(\int_a^b f(x)^n \,dx\right)^{1/n}$ converges when $n\to\infty$ and the limit is $\max_If$ with $I=[a, b]$.   This is my solution:  For  Weierstrass $f$ has maximum, $\exists \  \xi : f(\xi)=M$; and as $f$ is defined on $[a,b]$, $f$ is U.C., then:  $\forall \epsilon &gt;0  \ \exists \delta &gt;0: \forall x,y \in [a,b]: |x-y|&lt; \delta \Rightarrow |f(x)-f(y)|&lt; \epsilon$  Let $[a,b]=\bigcup_{k=1}^m I_k$, with $mis(I_k)&lt;\delta$, and $M_k=max_{\bar(I_k)}  f(x)$ $(\int_a^b f(x)^n \  dx)^{1/n}=(\sum_{k=1}^m M_h^n mis(I_k))^{1/n}=(M_1^n mis(I_1)+...+M^n mis(I_j)+...M_m^n mis(I_m))^{1/n}=$ =$M ((M_1/M)^n mis(I_1)+...+mis(I_j)+...+(M_m/M)^n mis(I_m))^{1/n}$  Then:  $(\int_a^b f(x)^n \  dx)^{1/n} \ge M$  $(\int_a^b f(x)^n \  dx)^{1/n} \le M(b-a)^{1/n}$  $\Rightarrow \exists \  \lim_n \ (\int_a^b f(x)^n \  dx)^{1/n}=M=\max_I \ f$
6371
I have seen that the method of "dominated convergence" implies that sometimes a limit cannot be brought inside of a sum, but is there a more basic way to understand this, preferably without using integrals?   A conceptual or simple answer would be helpful. Thank you!
107960
For $n\in N $ let   $a_n = \displaystyle \int_1^n \frac{\cos x}{x^2} dx$.   Prove, for $m \geq n \geq 1$, that $|a_m - a_n| \leq n^{-1}$ and deduce that $(a_n)$ converges.   By integration by parts, or otherwise, demonstrate the existence of $\displaystyle \lim_{n\to \infty} \int_1^n \frac{\sin x}{x} dx$.    I simplified $|a_m - a_n|$ into $\displaystyle \int_{n}^m \frac{\cos x}{x^2} dx$.  Now I'm stuck on how to make this less than or equal to $\frac{1}{n}$.
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?
76626
Prove that:  {$f_n$} converges unifromly to $f$ on $E$ as $n \rightarrow \infty$ if and only if $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$  I don't understand the approach for the Forward proof.  My approach is:  By definition, for $\epsilon &gt; 0$, there exist $N \in \mathbb{N}$ such that $|f_n(x) - f(x)| &lt; \epsilon$; hence, the smallest upper bound should also be less than the $\epsilon$ (i.e., ${sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon$). Thus, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  However, the solution uses $\frac{\epsilon}{2}$ as the following:     $|f_n(x)-f| \leq {sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon $      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2}$      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2} &lt; \epsilon$   Hence, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  Why?
111427
Previous related .  After I negated the definition of a convergent sequence, I ended up with the following mathematical statement:   $$\exists\ \epsilon &gt; 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n &gt; N : |x_n - l| \ge \epsilon$$  Is this correct? I'd like clarification...  Anyway, I'm now asked to use my negation of the definition of convergence to prove that $(a_n)=((-1)^nn)$ is divergent...  Am I right in assuming it's enough to prove that there exists at least one $\varepsilon$ such that $|(-1)^nn -\mathscr l|\ge\varepsilon$? Can I set $\mathscr l$ to be any real number I please?
1983
Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.   For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent.  Not to sure if this is correct, any help would be much appreciated, many thanks.
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?
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 $n\ge N \implies \left|\frac{1}{2n} - 1\right| = |\frac{1-2n}{2n}|$