How does one prove that Σ(2n)!!/(2n+1)!! diverges? The ratio test is inconclusive and I don't know how to use the comparison test and the limit comparison test in the situation.  Edit: I found an answer to my question .
121001
Prove convergence\divergence of the series: $$\sum_{n=1}^{\infty}\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$$  Here is what I have at the moment:  Method I  My first way uses a result that is related to  that we'll denote by $W_{n}$.  Also, we  may denote $\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}$ by $P_{n}$. Having noted these and taking a large value of $n$ we get: $$(P_{n})^2 =\frac{1}{W_{n} \cdot (2n+1)}\approx\frac{2}{\pi}\cdot \frac{1}{2n+1}$$ $$P_{n}\approx \sqrt {\frac{2}{\pi}} \cdot \frac{1}{\sqrt{2n+1}}$$   Further we have that: $$\lim_{n\to\infty}\sqrt {\frac{2}{\pi}} \cdot \frac{n}{\sqrt{2n+1}} \le \sum_{n=1}^{\infty} P_{n}$$  that obviously shows us that the series diverges.  Method II  The second way is to resort to the powerful  and firstly proceed with the ratio test: $$\lim_{n\to\infty} \frac{P_{n+1}}{P_{n}}=\frac{2n+1}{2n+2}=1$$ and according to the result, the ratio test is inconclusive.  Now, we apply Kummer's test and get: $$\lim_{n\to\infty} \frac{P_{n}}{P_{n+1}}n-(n+1)=\lim_{n\to\infty} -\frac{n+1}{2n+1}=-\frac{1}{2} \le 0$$ Since  $$\sum_{n=1}^{\infty} \frac{1}{n} \longrightarrow \infty$$ our series diverges and we're done.  On the site I've also found   with answers that can be applied for my question.  Since I've already have some answers for my question you may regard it as a recreational one and if you have a nice proof to share I'd be glad to receive it. I like this question very much and want to make up a collection with nice proofs for it. Thanks.
114562
Decide if the series converges and prove it using comparison test:   $\sum_{k=1}^{\infty}\frac{1}{2k+3}$   $$\frac{1}{2k+3} &lt;\frac{1}{2k}&lt;\frac{1}{k}$$  and since $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too.  Did I do everything correctly? What makes me feel a bit unsure is that there is no "$\leq$" sign anywhere and when I look up the comparison test on the internet, I don't see any "&lt;" or ">" in the definition.  Is it correct anyway?
97994
I know $4^{n}$ diverges, and $\frac{(n!)^2}{(2n)!}$ converges. I also think that their product diverges since the term of this series is increasing but I don't know how to prove it. I tried ratio test which is inconclusive, and some other tests.
8655
I am a little bit confused regarding the meaning of the phrase :" Root test is stronger than ratio test", and was hoping you will be able to help me figure it out. As far as I can see here:  The limit from the ratio test is greater or equal the limit from the root test . So, my first question is-  is there any example of a series $\Sigma a_n$ such that the limit from the ratio test is exactly 1 (i.e.- inconclusive), but the limit from the root test is less than 1? (i.e.- convergence can be proved by using the root test but not by using the ratio test ) If not, then is it correct that this phrase is the meaning of "stronger" is when the limit from the ratio test does not exist? (as in the classic example of a rearranged geometric series)  Hope you will be able to help.   THanks !   related posts:
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.
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.
123037
It is well known that $(n!)^2$ divides $(2n)!$.  Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$?  If yes or no, could you provide the details behind the argument?  As a quick test, I tried $n=3$ and $n=4$ which worked.  Thanks,  -Larry
81247
Does the following series converge or diverge? $$ \sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}} $$ The methods I have at my disposal are geometric and harmonic series, comparison test, limit comparison test, and the ratio test.
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.
88110
I am given the problem: Let $a_n&gt;0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.  We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect?
108931
I've been reading proofs of this theorem, and I was wondering why the following is true:  We know that $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ diverges.  I'm not sure how knowing this fact leads us to the next step:  For any $M &gt;0$ we can find primes $p_1,\cdots p_m$ such that $\left(\sum\limits_{i=1}^{N}\frac{1}{p_1^i}\right)\cdots \left(\sum\limits_{i=1}^{N}\frac{1}{p_m^i}\right)&gt;M.$  What I am saying is, given the opportunity, I would have never thought to look at a finite product of partial sums.  I understand why everything works after the above step.
115884
I ask for some help with this question:  Prove or provide counter example:  If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.  I tries this way:  If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.  There are 3 possible cases:  1) If $a_n &gt;0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}&lt;na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  2) If $a_n &gt;0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}&gt;a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.  3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.  Thanks.
16311
As seen in  question, the series $\sum\limits_{n=1}^{\infty} \dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)  However, more is true: not only are the terms greater than $1$, they blow up to $+\infty$! To see this, one can apply Stirling's approximation and get $\dfrac{n!e^n}{n^n} \sim \sqrt{2\pi n}$.  My question is: is there an elementary proof for $\lim_{n \to\infty}\dfrac{n!e^n}{n^n} = +\infty$?  The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+\infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.  (To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
114099
Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$  How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.  Regards,  Dennis
17721
How to test whether $\sum_{n=1}^{\infty }\frac{1}{n^{1+\frac{1}{n}}}$ diverges or converges? Can I apply ratio test? Thanks in advance.
38619
Problem :   Let $a$ be a positive number. Then $$\lim_{n \to \infty}\left[\frac{1}{a+n}+\frac{1}{2a+n}+\cdots +\frac{1}{na+n}\right]$$  Please suggest how to proceed in such limit problems, will be of great help thanks.
108212
I'm having a look at analysis right now, and I just thought up this question after reading about the comparison test.  Does there exist a "critical" infinite sum of real numbers (which is divergent) such that if $S_n$ is any infinite sum for which the terms in $S_n$ are all less than the terms in this critical sum for $n &gt; N_0$ for some finite $N_0$, then $S_n$ converges?   I know my terminology is somewhat incorrect.