clearly $\sum_{n=1}^{\infty} n $ is a divergent series  but there is a way to show that $\sum_{n=1}^{\infty} n =-\frac1{12}$ (not really meaningful to me)  My problems are,   1. can we show $\sum_{n=1}^{\infty} n $ is equal to another real number different from $-\frac1{12}$ ?  2. If there is a way, by using these results can we prove the series $\sum_{n=1}^{\infty} n $ is divergent?(using the method of contradiction)  thanks.
1958
$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) &gt; 1$.  Why should analytically continuing to $\zeta(-1)$ give the right answer?
27086
$$\sum_{n=1}^\infty \frac{i^n}{n}$$  I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
100736
Give an example of a convergent series $$\sum_{n=1}^{\infty}a_n$$ such that the series $$\sum_{n=1}^{\infty}a_{3n}$$ is divergent..  I cannot find find any kind of series... I am also be very thankful if you find a divergent series (changing the term convergent)...
13007
What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you!
113842
Given the series $A=\sum\limits_{n=1}^{+\infty}\frac{p_n}{p_{n+1}}$ and $B=\sum\limits_{n=1}^{+\infty}\frac{p_{2n}}{p_{2n+1}}$, where $p_n$ is the sequence where the nth number are the nth prime number, is it possible to determine whether they are converging or diverging?   I think there are some ways using the fact that $\sum\limits_{p\in\mathbb{P}}\frac{1}{p}$ diverges, but I'm not sure of how I can use it.
79
How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$  I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty  \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1}  \frac{1}{m^2+(N-m)^2}$$  $$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$ but it doesn't work.
85794
How to sum the series:  $$\sum _{ n=0 }^{ n=\infty  }{ \frac { 1 }{ { 2 }^{ { 2 }^{ n } } }  }$$   PS: Just a hint would suffice.
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
57775
What is the value of  $$\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}$$  It's clearly convergent but is it possible to calculate the sum?
15494
Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent.  I've tried applying all the basic tests to no avail. I need to find out what the "trick" for this particular series is.
38229
Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?  Thanks!
1163
Does this series converge or diverge? $$\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$$   my thought is that, I can break it down to $$\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)$$ then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges.  Any help would be appreciated!
54251
Show that the following sum is divergent   $$\sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)$$   I thought to do this using Taylor series using the fact that $$ \ln\left(1+\frac1n\right)=\frac1n+O\left(\frac1{n^2}\right) $$ Which then makes it clear that $$ \sum_{n=1}^{\infty}\ln\left(1+\frac1n\right)\sim \sum_{n=1}^{\infty}\frac1n\longrightarrow \infty $$ But I feel like I overcomplicated the problem and would be interested to see some other solutions. Also, would taylor series be the way you would see that this diverges if you were not told?
76052
I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on  and found this bewildering formula,   $$1 + 2 + 3 + \dots  = - \frac1{12}$$  The article also says that "Ramanujan summation of a divergent series is not a sum in the traditional sense". I am wondering then what this summation actually implies?  Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand?
48067
I can clearly see that $\dfrac{n!}{n^n}\to 0$ when $n\to\infty$. But how do I know if the sum $$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$ is convergent or not? I know this might be basic, but thank you if anyone can help me.
15598
Classify  $$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$ as absolutely convergent, conditionally convergent or divergent.  Is it, $$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and $$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.  So, $$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$ is conditionally convergence?  I'm not sure if I'm doing right or not. Could you guide me?  Thanks in advance! :)
77115
Let $a_n=1-2^{\frac{-1}{n}}$, $b_n=\frac{1}{n^p}$. Does $\sum_{n=1}^{\infty}a_n$ converge or diverge when compared to $\sum_{n=1}^{\infty}b_n$ if $p &lt; 1$? My answer to this is that it diverges, since $b_n$ diverges if $p&lt;1$, and by a comparison test $a_n$ does as well. Is this correct? Thank you.