I guess it is kind of obvious that it diverges. But what divergent series do I compare it to?
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.
60753
The series $$\sum_{p\;\text{prime}}\frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots$$ diverges as is well known. How about the following? $$\sum_{p\;\text{prime}}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{11!}+\cdots$$ Its convergence is easily obtained by comparing with the series $e=\sum_{n=0}^\infty\frac{1}{n!}$. Numerically, we get $$\sum_{p\;\text{prime}}\frac{1}{p!}\simeq 0.675198$$ Can we find the specific value?
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.
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! :)
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.
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.
106960
I read this series around 5-10 years ago. There were two parallel universes, ours and the universe of demons, and much of the series revolved around attacks in which demons created gateways to cross into our universe to kill people and wreak havoc, and had to be stopped by human magicians. Three youngsters (one guy from modern times, one from a few decades ago, and a girl from ancient Celtic times, all with funny names) end up being the key to the war when it's revealed that they are the three components of a mythical superweapon. These three are the only point-of-view characters in the series, each book being told from the PoV of just one of them. At the very end, they   destroy our entire universe so that, after killing all the demon masters, they can rebuild it and make sure everything happens as before up until the very last moment.  As they begin recreating the universe, the current PoV character (the one from modern times) says something like the following, the very last lines of the series. (sic on the quotation marks)  &quot;In the beginning, there was nothing but darkness. Then [name] said, 'Let there be light!'&quot; And there was light. Coolio!
34268
The alternating series test (AST) says, briefly, that if   $a_k&gt;0$ $a_k \geq a_{k+1}$, and $a_k \to 0$ as $k \to \infty$   then $\sum_k (-1)^k a_k$ converges.  This seems to be a one-way test (that is, if an alternating series fails the test, we don't know that it diverges).   document says so explicitly. It then gives an example of an alternating series which fails the AST. That doesn't prove the series is divergent (but it turns out to be divergent, anyway, by $n$th term test).  Is there a convergent, alternating series that fails the AST?  Of course, if the series fails condition 3, then it also fails the $n$th term test, and must diverge. And if it fails the first condition, then it's not strictly alternating, anyway.  So it must be a series that is not getting smaller (condition 2), but still converges.
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.
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
68121
Does the following series converge or diverge?  $\sum_{n=1}^{\infty} a_n$ where  $a_n = \frac 1 b_n$, and $(b_n)_n$ is the subsequence of $(n)_n$ whose terms do not have a $9$ in their decimal notation.  I have no idea how the sequence $b_n$ even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what $b_n$ is.
104617
Wolfram alpha stated that this series converges to $0,$ but when I entered this answer in my homework it says it is incorrect, therefore I wanted to know whether this series diverges or converges.
68700
I came across this when learning about sequences and series. No one proved it to me tho! Is there a link anywhere? Or would it be beyond what I know? (First term of analysis)
109414
If we are given any sequence of real numbers $\{a_n\}$ diverges, so $a_n \rightarrow + \infty$, how can we find a sequence $\{b_n\}$ such that $\sum |b_n|$ converges but $\sum |a_n||b_n|$ diverges?  I want to use this fact in another problem but don't immediately see how to prove it.
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?
48636
I read it as a paperback around 2009ish. There are at least two books in the series. In the first book, we learn that the protagonist lives up in the mountains. Her people are distrusted by the people in the plains, and I think have either dark or olive skin. She leaves home to learn magic, something I think is rare in the world she lives in. When she arrives, there's a crazy power-hungry man at her destination and she has to hide who she is because otherwise he'd have her killed to prevent her from surpassing him. At some point, she winds up in a set of chambers set into the mountain where she finds a pool of water and carvings about dragons. Somehow, this leads into her learning to become a dragon, I think perhaps after meeting one. I want to say that the sequel got into her having dreams that turn out to be partly prophetic and partly communication.  One of the reasons the book sticks out in my memory is that, despite the mentions of the protagonists dusky skin, the cover depicts a fair-skinned blonde girl, I think holding a glowing globe of light.
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?