Show that $1 + 1/2 + 1/3 + .... + 1/n - \ln (n)$ converges. I tried to prove it using the comparison test. First I made it into a sum of 1 + $\sum_{2}^{n} \frac{1}{m} -\ln(\frac{m}{m-1})$ and tried to do comparison with $\frac{1}{m^2}$. From $|\frac{1}{m} - \ln(\frac{m}{m-1})| &lt; \frac{1}{m^2}$, I managed to prove  $\frac{1}{m} - \ln(\frac{m}{m-1}) &lt; \frac{1}{m^2}$ but I am unable to show that $-\frac{1}{m} + \ln(\frac{m}{m-1}) &lt; \frac{1}{m^2}$. I am wondering how to prove the last inequality if its possible.
47677
I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges. At first I want to know if my answer is OK.  My try: $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$     $ = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ Now we prove that the last sum converges by the comparison test: $\frac{1}{k}-\log(\frac{k+1}{k}) &lt; \frac{1}{k^2} \Leftrightarrow k&lt;k^2\log(\frac{k+1}{k})+1$ which surely holds for $k\geqslant 1$      As $ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $ \Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ converges and we name this limit $\gamma$ q.e.d
57252
How to prove that the series $$\sum_{n\geq 2}\log\left(1-\frac{1}{n^2}\right)$$ is convergent? What about finding the sum?  My attempt: $$\ln (1-1/n^2)= \ln(n-1) -2\ln n + \ln(n+1)$$ The term in the log is decreasing so I get the feeling that the series converges but confused how to prove this and also find the sum.
76880
How to test the convergence for $$\sum _{n=3}^{\infty }\:\left(-\frac{1}{\sqrt{\ln\left(n\right)}}\right)^{n+1}$$ Can I use the comparison test for this one? For example: $n≥\ln(n)$ $\sqrt{n}\ge \sqrt{\ln\left(n\right)}$ $-\sqrt{n}\le -\ln\left(n\right)$ $\frac{1}{-\sqrt{n}}\ge \frac{1}{-\sqrt{\ln\left(n\right)}}$ and then I use the the p series for p = 1 and say that this diverges? or can I only use this inequality for convergent? Thanks :)
112564
$|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here.
70654
$$\sum_{k=1}^{\infty} \frac{1}{k^{\ln k}}.$$  I have some problems with this sum, I have tried most of the common methods, such as the ratio/root test, integral test, etc. I do believe it looks like a problem easily solvable by a direct comparison with some p-series, however, I do not know how to formulate the inequality.   I did try using the common fact that $\ln(x) \ll x^p$ for large $x$ but that of course only bounds it from below with a convergent series saying nothing about ours. Also I tried rewriting the 'summand' as $e^{\ln(k^{\ln k})}=e^{\ln k \cdot \ln k}$ but this did not really yield anything.   Any answer or hint would be greatly appreciated.
2562
I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:  Monotonic:  The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$ $$a_1=1\leq 1+\frac{1}{2^2}=a_2$$  Need to show that $a_{n+1}\leq a_{n+2}$ $$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$  Thus the sequence is monotone and increasing.   Boundedness:  Since the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.   Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?  Thanks so much in advance!
121784
I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this.   $$\frac 1{x+1}&lt;\ln\left(\frac{x+1}x\right)&lt;0.5\left(\frac 1x+\frac 1{x+1}\right)$$
89005
I have to prove that $$\lim_{n\to\infty} \left(1+n^2\right)^{\frac1n}$$ converges and give its limit. What I did was the following: $$\lim_{n\to\infty} \left(1+n^2\right)^\frac{1}{n} = \lim_{n\to\infty}(n^2)^{\frac1n}\lim_{n\to\infty}\left(\frac1{n^2} + 1\right)^\frac1n= \lim_{n\to\infty}(n^2)^{\frac1n}$$ Now that is where I am stuck - how to I continue with $\lim_{n\rightarrow\infty}(n^2)^{\frac{1}{n}}$? I really don't have any ideas left... Thank you very much! FunkyPeanut
99381
I'm trying to prove this sequence converges:  $\sum_{n=1}^{\infty}  \frac{1}{n\ln^2(n)}$  I noticed that this is continuous function which its derivative is always less than $0$ for $ x \gt 1 $, so I tried to do the integral test but with no success...   Some help?  Thanks!
23618
If $a_n \ge 0 $ for all n, prove that $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty {a_n\over 1+a_n}$ converges.   Here is my attempt!  => Suppose that $\epsilon \ge 0$ is given and $a_n$ converges, then  for all $N\le n \le m$$$\sum_{k=n+1}^m a_k \lt \epsilon.$$ let $b_n={a_n\over 1+a_n}$ and Notice that $b_n \lt a_n$ and   Then, applying the comparison test  $$|\sum_{k=n+1}^m b_k| \le \sum_{k=n+1}^m |b_k |\le\sum_{k=n+1}^m a_k\lt\epsilon$$Hence,${a_n\over 1+a_n}$ converges.  let me know if it is worng and I don't know how to start for the other direction. Any help is appreciated!
86369
Check whether $$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{\left(m+n\right)^2}$$ converges or NOT?  My Try:- $\sum _{m=1}^{\infty }\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(m+n\right)^2}=\lim_{j\to \infty}\sum _{m=1}^{j }\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(m+n\right)^2}=\lim_{j\to \infty}(\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(1+n\right)^2}+ \lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(2+n\right)^2}+ \lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(3+n\right)^2}+...\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(j+n\right)^2})\ge \lim_{j\to \infty}(j\lim_{i\to \infty} \sum _{n=1}^{i}\frac{1}{\left(j+n\right)^2})$  How do I complete the conclusion?
96520
Kind of stuck in this one. I've tried substracting $$\ln(1+(n+1))\leq\sum_{i=1}^{n}\frac{1}{i}+\frac{1}{n+1}\leq1+\ln(n+1)$$ at the original inequality and applying properties of the logarithms, but then I don't know how to keep going :/
58157
Show that $$1+\frac {1}{4} \bigg(1+\frac {1}{4}\bigg) +\frac {1}{9} \bigg(1+\frac {1}{4} +\frac {1}{9}\bigg)+.....$$  converges.  Can you find the exact value of the sum.  My effort:  I have proved the convergence with comparing to $$\bigg(\sum _1^\infty \frac {1}{n^2}\bigg)^2$$  I have not figure out the exact sum.  Any suggestions??
34945
I am sorry that my question is not whether $\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}$ convergent or not.  My question is how to compute the sum of series:  $$\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},$$ It is a question in a textbook, and I have worked on it for days, but fail.
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?
76726
$$\frac{1}{n+1}&lt; \log(1+ 1/n)$$  Any ideas? I tried estimating the difference between $1/n$ and the logarithm and comparing with $1/n-1/(n+1)$ but I miserably failed.
16407
How do I prove that if $|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|$ and $\sum |b_n|$ is convergent, then $\sum |a_n|$ is convergent? I know that if I can get $|b_{n+1}/{b_n}| &lt; 1$ then I can use the ratio test to say $\sum a_n$ is convergent, but I don't know if this the proper way to approach the problem.