I find this series to be a bit troublesum. I can't find a proper method that works. Root test doen't seem to be useful. Ratiotest got ugly. Integral test seems to force me to integrate a function $x^{1/x},$ which doesn't have any elementary primitive.  I can't find an apropriate limit to compare it to and do a limit test comparison.
46964
A student was recently asked this question by his instructor:  $$\sum_{n=1}^\infty{\left(\sqrt[n]{n}-1\right)}$$  Converge or diverge?  I feel a little dumb for not being able to answer it. The following tests fail to prove convergence or divergence:  nth term test for divergence (limit is 0), ratio test (limit is 1), root test (see ratio test), limit comparison with $\sqrt[n]{n}$ (not sure why I thought that'd work)  Something I did try was using the fact that  $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1})\;,$$  to rewrite $\sqrt[n]{n}-1$ as  $$\frac{n-1}{n^{1-1/n}+n^{1-2/n}+\cdots+n^{2/n}+n^{1/n}+1}\;.$$  However, I'm not sure what to compare this to. According to wolfram alpha this series "diverges by the comparison test", but comparison to what? There is a similar problem in Baby Rudin, but for $(\sqrt[n]{n}-1)^n$, and a simple nth root test resolves that series [convergence] in a hurry. Any ideas? Have any of you encountered a similar looking series before? Thanks.
56578
$$ \sum_{i=0}^\infty \frac{(-1)^n}{n} $$  This alternating series fails the p-series test because the exponent of n = 1.  Yet it seems to pass the alternating series test.   1 - $a_n$ must be positive. True.  2 - Terms must be decreasing. $\frac{d}{dn} 1/n = -n^{-2}$, which is &lt; 1. True.  3 - $ \lim_{n\rightarrow\infty} 1/n = 0  $ True.  $(-1)^n/n$ is clearly a divergent series, so why does it pass the AST?
96253
Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by $\frac{1}{n^2}$ converges.
11488
so I used the root test, but i'm not quite sure if i'm allowed to. I think im performing the operations correctly,a and i keep ending up with $(1)^{\infty}$. So really my question is am i performing the operations wrong or do i have to use a different test?
34891
I need to find out whether the following series converges or diverges. If it converges I also have to discover whether it is an absolute convergence or a conditional one. $$\sum_{k=1}^{\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^{2k}$$ Root test did not help me much since $\lim_{k\rightarrow\infty}\left(\frac{1+\cos k}{2+\cos k}\right)^2$ may equal anything from $0$ to $4$. Comparison tests also did not work well. I guess this series is convergent, though. But I still have no idea how to prove it.
88183
This question is realy hard for me and I'd like to get some advises:  $$f(x) =   \begin{cases}      x^{\alpha}(1-\cos(1-x))^{\beta} &amp;  x\in (0,1)\\[6px]     \dfrac{1}{x^{\alpha}+x^{\beta}}  &amp;   x\ge 1   \end{cases}$$  I need to find the values of $\alpha$ and $\beta$ that make $\int_{0}^{\infty}f(x)\,dx$ converge.  Now I think I need to split the integral to $\int_{0}^{1}f(x)\,dx$ and  $\int_{1}^{\infty}f(x)\,dx$ and check the converge for each integral. But I'm stuck here.
21194
So my teacher gave a pretty tough (for me) problem in class today: Does the series $$ \sum_{n=1}^{\infty}\frac {\sin{\frac 1 n}} {\sqrt n} $$ converge or diverge?  So far I've thought of trying the Comparison Test and Ratio Test but couldn't really get anywhere. Intuitively I think it should converge, but how do I actually show this?
15740
Ratio and root tests won't help. And I can't use the comparison test because $ |a_n| $ is not necessarily smaller than $ n^2a_n $.   Can I use limits? We know:    $\lim\limits_{n \to \infty} a_n = 0 $ $\lim\limits_{n \to \infty} |a_n| \ne 0 $   And we need to prove:    $\lim\limits_{n \to \infty} n^2a_n \ne 0 $   Any ideas/hints?
116928
I was experimenting with various integration techniques, and I stumbled upon the integral:  $$ \int{\frac{\tan{x}}{x} dx} $$  I tried using a number of methods, but I could not solve it. Checking on WolframAlpha, I found that the integral does not have an elementary antiderivative. However, I am not satisfied with simply assuming it's non-elementary if I have difficulty in solving it.   My question: Is there a test to determine if an antiderivative is elementary or not?
114346
I am trying to study the convergence of the series:  $$\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$$ First I have failed to use the D'Alembert test  $$ \lim_{n\to \infty} \frac{(\ln (n+1))^{\ln (n+1)}}{(\ln n)^{\ln n}} =1.$$ So I can Conclude anything from here.  I have also tried the integral Comparaison test but it was not soo clear.
69338
The following is a thought on the proof of the integral test. I am looking forward to knowing if it provides a correct understanding.     For $n&gt;N$ we can write, $\sum_{n=N}^{\infty} a_n = \sum_{k=N}^{\infty} f(k)$.   We can observe, $\sum_{k=N}^{b-1} f(k) \Delta x &gt; \int_{N}^{b}f(x)dx$, where the L.H.S of the inequality is the upper Riemann sum, $b-1&gt;N$ and $\Delta x$ is a constant.   1) $\lim_{b\rightarrow \infty} \int_{N} ^{b}f(x)dx$ diverges.   It is clear that $\sum_{k=N}^{\infty} f(k) \Delta x$ diverges. Since $\Delta x$ is a constant, we can write,  $ \Delta x \sum_{k=N}^{\infty} f(k)$. This tells us that the series diverges.   We can observe that the lower sum is less than the finite integral. The series corresponding to the lower sum is $\sum_{k=N+1}^{b} f(k)$. As $b\rightarrow \infty$, the series can be written as, $\sum_{k=N}^{\infty} f(k) - f(N)$. This diverges because the infinite sum diverges.   2) $\lim_{b\rightarrow \infty} \int_{N}^{b}f(x) dx$ converges.   Here, the lower sum converges because it is less than the finite integral. We can say, $\Delta x\sum_{k=N+1}^{b} f(k)$ converges and so does the series.   The series corresponding to the upper sum is $\sum_{N}^{\infty} f(k)$. This can be written as $\sum_{N+1}^{\infty} f(k)+f(N)$. This converges because the series converges.   This ends the proof.
76881
$$\int^\infty_1\frac{1}{x^{1+1/x}}dx$$  I'm preparing for exams. I would also like to know what are some commonly used methods to show an improper integral diverges?
84707
Find whether the integral converges or diverges. $$\int_{0}^{1}\frac{\log(x)}{1-x^2}dx$$  I simplified it to $$\int_{0}^{1}\frac{\log(x)}{(1-x)(1+x)}dx$$  Here I have $2$ "bad" bounds (both $0$ and $1$).  I think, to prove such integral converges, I should find something that is equivalent to or less than $\log(x)$ so I can exchange them, maybe $\sqrt{x}$, so it may cancel out the fraction, and I can compare it to $$\int_{0}^{1}\frac{dx}{x^p}$$  Also I had an idea to substitute $x$ with $u+1$ so I can use $\log(1+u) \sim  u$ when $u \rightarrow 0$. I'm allowed(not familiar with anything else) to use direct comparision test, limit comparision test, Cauchy test and Abel-Dirichlet's test.   Any hints are much appreciated, I just want to get the big picture.   Thank you.
25486
Task: find all values of the parameter, such that integral converges. $$\int_0^{+\infty} \frac{dx}{1+x^a \sin^2x}$$ I tried a lot and I used Cauchy and Weierstrass method but it was useless. And now I know that I must use $$\sum_n \int_{\pi n}^{\pi(n+1)}$$
12551
If $x&gt;0$, find the set of all values of $x$ such that series is convergent$$\sum_{n=1}^{\infty} x^{\ln{n}}$$   My attempt:- I used Ratio test for finding the set of all values of $x$ such that series is convergent.   $$\lim_{x\to\infty}\frac{x^{\ln{n+1}}}{x^{\ln{n}}}$$  $$=\lim_{x\to\infty}x^{\ln\frac{n+1}{n}}$$ This quantity must be less than one for getting a convergent series, I am not able to judge. Can you please help me to find the interval of convergence?
82381
Is the following series convergent?  $$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$  I treid the Ratio and Root tests, but both of them failed.
55937
I have to decide whether the Series $\sum_{n=1}^\infty \frac{n^2}{2^n}$ Converge or Diverge only by using the Comparison test or by $\frac{1}{n^p}$ test.  What I noticed is that it converges by the De'lamber test ($\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$).  But I'm having trouble finding a way proving it with the provided tests.  Would appreciate your advice