Say I have this series:  $$\sum_{n=1}^{\infty} \frac{n+3^n}{n+2^n}$$  I think I can compare it to $$b = \frac{3}{2}^n$$ which is a geometric series that diverges.  But where do I go from here? The issue is that the numerator and denominator both increase so I can't say that the original series diverges or not by Comparison Test. What else can I do?
33274
The sum is $$\sum_{n=1}^{\infty} \frac{n+2^n}{n+3^n}$$ Is this convergent or divergent? I tried to use the divergent test but the test fail because $a_n = (n+2^n)/(n+3^n) = 0 $ as $n$ goes to infinity. Could someone point me to the right direction?  Thanks
50422
I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges.   I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive.   So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence.
123584
Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1". Thanks a lot.
91354
Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges   I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.  Any hints?   Thank you!
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
24869
So I've been thinking through some test cases. If $a_n = n$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ is the harmonic series which diverges. And if $a_n = \sum_{k=1}^n 1/k$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverges like $\sum_n 1/(n \log n)$. So that got me thinking, if $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ necessarily diverge?
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!
40706
Let $a_n$ be a sequence. $\lim_{n\to\infty} a_n =0 $ then the series: $\sum_{n=1}^\infty a_n, \ \sum_{n=1}^\infty (a_n+a_{n+1}) $ converge and diverge together.  It doesn't seem right, there are no convergence tests that fit this question yet every counterexample I try does not seem to work. Is this how I should see the sum? $\sum_{n=1}^\infty (a_n+a_{n+1})=2\sum_{n=1}^\infty a_n +a_{n+1}$ If so then there's no real difference between the two series since you multiply it by a constant and add just one more value.
23900
Per the title, if $\sum_{1}^{\infty}(a_n)^3$ diverges, does this imply that $\sum_{1}^{\infty}(a_n)$ diverges?   I'd appreciate hints (!) for dealing with this excercise.  EDIT Per the contrapositive, it is not given that $a_n$ converges absolutely, or that it is nonnegative for all $n$.  Thank you!
38303
As the title says, given a series $b_n &gt; 0$, where $\sum_{n=1}^\infty b_n$ is divergent:  Show that the series $$\sum_{n=1}^\infty \frac{b_n}{1+b_n}$$ is also divergent.  So I've defined the series as $\sum_{n=1}^\infty u_n v_n$ where $u_n = b_n$ and $v_n = \frac {1}{1+b_n}$.   I'm sure there was some sort of convergence test which stipulated that for a product of two series, if one diverges the entire thing does, however I might be mistaken. If so, how do I properly prove this?
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.
95832
$\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}} = \infty$. Is there a comparison that works well to prove this?
50138
Could someone give me an example of two convergent series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ such that $\sum_{n=0}^\infty a_nb_n$ diverges?
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
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)...
78477
It is obvious that   $\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}(2n-1)}{2^{n-1}}$  is alternating series, therefore, before we could talk about finding the sum of the series, we have to check if it's unconditionally convergent, if we look at $a_n=\frac{2n-1}{2^{n-1}}$ it is obvious that we could use d'Alembert's test and easily prove that this is unconditionally convergent series. So now, we should find the sum:  $S_n=\sum_{k=1}^{\infty} \frac{(-1)^{n-1}(2n-1)}{2^{n-1}}$  because of the denominator, it really looks like geometric series, but i could't find a way to simplify it in order to determine the sum. Any ideas?
44494
$$\sum_{n=1}^\infty \frac 1{n!}$$   I think we may compare it with $\frac{1}{n^n}$, and say that $\frac{1}{n!}&gt;\frac{1}{n^n}$. But I am not sure of the convergence or divergence of $\frac{1}{n^n}$ itself, so this turns out to be useless.  What other way is there?