Question :    For each $n \in \mathbb N$, Define : $\gamma_n=1+\frac{1}{2}+\dots+\frac{1}{n}-\int_1^n \frac{1}{x}dx$    Prove that $\{\gamma_n\}$ is convergent.      Note 1 :  I know that $\sum_1^\infty \frac{1}{k}$ diverges. I can guess that this sequence is convergent to $0$ because of the similar terms inside the integration and outside of it.  The problem is that the integration is continuous. I mean that $x$ is not just integers.  What can we do with the values left?! ( For example, $\frac{1}{2.5}$ ) Also, Notice that this problem should be solved with some elementary tools.  ( It's from a part of my book which is even before any discussion about the Fundamental theorem of calculus. )   Note 2 : I'm not good at integration and i'm learning it. Any good details can help me.    Thanks in advance.
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Problem Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent. One Proof This proof is based on the following inequality  $$\frac{1}{n+1}&lt;\ln \left(1+\frac{1}{n}\right)&lt;\frac{1}{n}$$  where $n=1,2,\cdots$, which will be used repeatedly. On one hand, we obtain that $$\ln 2-\ln 1&lt;1,~~\ln 3-\ln 2&lt;\frac{1}{2},~~\ln 4-\ln 3&lt;\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n&lt;\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)&lt;1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)&gt;\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is bounded below. On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists. Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.
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Question is to check if : $\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}&gt;\frac{3}{5}$   the problem is that i am sure that this series$\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}$ is convergent (if it is not then i would be very happy as i will then be done.)  I would be thankful if someone can help me out with solving this.
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Showing the sequence converges $a_{1}=\frac{1}{2}$,  $a_{n+1}=\frac{1}{2+a_{n}}$.  I already know that if $(a_{n})$ converges then it does to  $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.  EDIT I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
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Determine this series converge or diverge, and if it converges, find its value.  $$\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}$$  This was too hard for me, as unboundedness of $\tan{x}$ at infinite number of poles make harder to guess the behavior of value of tangent function at positive integers.  Thanks.
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For $n=1,2,3,\dots,$ and $|x| &lt; 1$ I need to prove that $\frac{x}{1+nx^2}$ converges uniformly to zero function. How ?. For $|x| &gt; 1$ it is easy.
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Let $c&gt;0$, $a_{1} = 1$, and $$a_{n+1} =\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)$$  I need to:   Show that $a_{n}$ is defined for every $n\geq 1$ Show that this sequence is convergent. Find its limit.   I proved the first part by showing by induction that this sequence is positive for every $n$. To show that this sequence is convergent I'm thinking of showing that this sequence is a Cauchy series, yet can't figure out how.  For the third part I'm clueless at the moment.
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This comes from the comments section of this question , credits .  The statement is  $$\sum_{n=1}^\infty\left(e-\sum_{k=0}^n\frac1{k!}\right)=1$$  This looks really interesting, so I was wondering if anyone has any ideas/suggestions or even better the whole proof?  Disclaimer: I added that link specifically because this might look like homework, leading to "What have you tried?". So the disclaimer is, I actually haven't tried anything yet, because yes, asking the community is just easier. It would be great if someone could answer this though since I, personally at least, think that this equation is very interesting and adds value to any math enthusiast stumbling upon it here someday. :) I will myself be trying it meanwhile too though(it really is interesting) , and add and accept my solution if I am successful and there are no answers yet.     I tried  for $\sum_{k=0}^n\frac1{k!}$ and it yields  $$\sum_{k=0}^n\frac1{k!} = \frac{e\Gamma (n+1,1)}{\Gamma(n+1)}$$  where $\Gamma(a,r)$ is the incomplete gamma function and $\Gamma(a)$ is the euler gamma function. . Might be of help, might not. Just thought it'd be worth adding.
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I need to prove that:  $$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$$  is convergent, but not absolutely convergent.  I tried the ratio test:  $$\frac{a_{n+1}}{a_n} = -\frac{\log\left(1+\frac{1}{n+1}\right)}{\log\left(1+\frac{1}{n}\right)} = -\log\left({\frac{1}{n+1}-\frac{1}{n}}\right)$$  I know that the thing inside the $\log$ converges to $1$, so $-\log$ converges to $0$? This is not right, I cannot conclude that this series is divergent.  Also, for the sequence without the $(-1)^{n+1}$ it would give $0$ too.
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I am trying to prove that   $\lim_{n\to\infty} (\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) } = e^γ$ where $γ$ the Euler-Mascheroni constant. We know that that:  $\lim_{n\to\infty} \frac{\sum^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$  By approximating the sum with integrals $$ \ln(n+1)=\int_1^{n+1}\frac1x\mathrm dx\le\sum_{k=1}^n\frac1k\le 1+\int_1^n\frac1x\mathrm dx=1+\ln n $$ since $1/k$ is decreasing for $k\ge 1$. We have that $$ \frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln(1+\frac1n)}{\ln n}\to1 $$ as $n\to\infty$ and we obtain the result. But I can't calculate $\lim_{n\to\infty} (\frac{\sum^n_{k=1} \frac{1}{k}} { \ln(n) })^{ \ln(n) } = e^γ.$ Any help?
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The convergenceradius is $1$ that means if $|z|= 1$ then the convergencebehaviour depends on $d$.   I have several questions about this issue:  1.Why is $\sum_{n=0}^{\infty}\frac{1}{n^d}$ absolutely convergent for $d&gt;1$ and $|z|=1$  2.Why is the implication true:  series absolutely divergent $\Rightarrow$ series is divergent  3.Why is $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ divergent for $d\leq 0$  and $|z|=1$  4.Why does the convergencebehaviour depends on $z\in\{a\in\mathbb{C}||a|=1\}$ for $0&lt;d\leq 1$?  I know that if $d$ is for example $1$ then  $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ converges absolutely for z = $-1$ and diverges for z = $1$. But the  statement above says we can go with $d$ as close to 0 as we want and we will still find a $z$ with $|z|=1$ for which  $\sum_{n=0}^{\infty}\frac{1}{n^d}z^n$ converges can somebody please show me the proof for all those questions I have?  Thank you!
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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?
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Please help with my homework question:  Define $\{a_n\}_{n=1}^{\infty}$, $\{b_n\}_{n=1}^{\infty}$ as:  $b_{n+1}=\frac{a_n+b_n}{2}, a_{n+1}=\frac{a_n^2+b_n^2}{a_n+b_n}, 0&lt;b_1&lt;a_1$  Prove the limits converge and that $\lim \limits_{n \to \infty}a_n = \lim \limits_{n \to \infty}b_n $.  I have tried to show that they are monotone and bounded, tried looking for a property of arithmetic mean (related to $b_n$) but I'm completely stuck as none of these yielded anything useful.  Any help would be appreciated
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I have tried a little bit which as follows- Since $f(x_n)\in[0,1]$, $\{f(x_n)\}$ has a convergent subsequence say $y_n=f(x_{r_n})\ \forall n\in\Bbb{N}$ Let, $\lim y_n=l\implies \lim \frac{y_1+y_2+\cdots+y_n}{n}=l\implies \lim \frac{f(x_{r_1})+f(x_{r_2})+\cdots+f(x_{r_n})}{n}=l$ But I am getting no idea to proceed and prove the convergence of $\{x_n\}$. I have also tried to prove the sequence to be cauchy which goes- $x_{m+1}-x_{n+1}={\sum_{i=1}^m f(x_i)\over m}-{\sum_{i=1}^n f(x_i)\over n}\le {\sum_{i=1}^m f(x_i)\over n}-{\sum_{i=1}^n f(x_i)\over n}$ (since $m\ge n)$ $\implies |x_{m+1}-x_{n+1}|\le \frac{|f(x_m)|+|f(x_{m-1})+\cdots+|f(x_{n+1})|}{n}\le\frac{m-n}{n}$ (since $f([0,1])\subseteq [0,1])$ Now, what will I get if I tend $m,n\to\infty$? But I need to do something more in the 2nd case since I have nowhere used the continuity of $f$.  Can anybody give an idea to prove it? Thanks for assistance in advance.
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Let $t_n= \frac{1}{n} [1+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}]$, n=1,2...  Then I am asked whether series$ \sum t_n $ converge or diverge. Also whether sequence $ t_n $ converge to zero or not.  I tried About sequence $t_n$, by Cauchy's 1st theorem, that since 1/$\sqrt{n} $ converges to zero so does t_n. But the seriex part is doubtful to me.?
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Show that $$\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent.   $$\int_{0}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent by comparison test with $\int_{0}^{\infty} \dfrac{1}{x^4} dx$.  How do i show that $$\int_{-\infty}^{0} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent?
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I have a question in Analysis. Knowing that $x_{2n}$, $x_{2n-1}$, $x_{3n}$ converge, how can I show that $x_{n}$ converges?
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Please help me solve this. Please try to give some details also.  Let $a_n=\sum_{k=1}^n  \frac{n}{n^2+k}\ $for $ n \in\mathbb{N}$ then the sequence $\{a_n\}$ is (choose the correct option):     Convergent   Bounded but not convergent   Diverges to infinity   Neither bounded nor diverges to infinity     My thoughts : I tried doing it by testing if it is monotonic with help of $a_{n+1} -a_n$ and checking if its +ve or -ve but, the the terms dont simplyfy by cancelling each other...