Prove that $\sum_{k=1}^\infty \sqrt{k+1}-\sqrt{k}$ diverges. I know that this diverges but I don't know how to prove it rigorously. Should I use the comparison theorem? I know that $\sqrt{k+1}-\sqrt{k}&gt;-\sqrt{k}$ so I just need to show that $\sum_{k=1}^\infty -\sqrt{k}$ diverges. But that is where I am stuck.
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.
55381
How would you show that the series $$\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$$ diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avail: The ratio test is inconclusive, comparing with something "smaller" which I know diverges is proving difficult. If possible, I'd like to show this without using anything fancy (such as Stirling's approximation).  Can you do this with a "basic" comparison?
53946
Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$  What method to use in order to find the closed form of summation $\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}$?
12931
$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$  I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.  Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?
82888
Found the alternative form: $\sum _{k=1}^\infty ((1-\frac{1}{k})^{k})^k$. Tried various criteria, no luck so far.
34647
Does the following series converge or diverge? I would like to see a demonstration.  $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}}. $$  I can see that: $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2}}(n)}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2n}}(n)}}. $$
52332
I have a question which is giving me a hard time.    I want to show that $$ \lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x$$ given that $\lim_{n\to \infty} x_n= x$.
105500
$\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$ $\\$  I tried a comparison test, Leibniz, Cauchy, Cauchy-integral and D'Alembert criteria but nothing gives me the result. Any hints would be helpful!
56471
Give either a justification or a counterexample for each of the following statements about real series. $\quad(a)$ If $ka_k\to0$ as $k\to\infty$, then $\sum a_k$ converges. $\quad(b)$ If $\lim\limits_{k\to\infty}\frac{a_{k+1}}{a_k}$ exists and equals $L&gt;1$, then $\sum a_k$ diverges. $\quad(c)$ If $\sum a_k$ converges and $\lim\limits_{k\to\infty}\frac{a_k}{b_k}=1$, then $\sum b_k$ converges.  I think all propositions are false.  But struggling to find counterexamples.  Is there some sort of systematic way of constructing counterexamples for these sorts of series?  Can someone explain their thinking process in giving counter examples for the above?
66694
I'm currently tackling the following infinite series in order to determine if it converges or not:  $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n}$$  What I've done is rearranged this to make use of the comparison test, shown below:  $$= \frac{\sqrt{n+1}-\sqrt{n}}{n} * (\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$  $$= \frac{(n+1)-(n)}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$  $$= \frac{1}{n^{3/2}+\sqrt{n+1}} &lt; \frac{1}{\sqrt{n+1}} &lt; \frac{1}{\sqrt{n}} $$  However, $\frac{1}{n}$ is smaller than $\frac{1}{\sqrt{n}}$, therefore:  $$  \frac{1}{n}   &lt; \frac{1}{\sqrt{n}}$$   This implies divergence by the comparison test due to the fact $\frac{1}{n}$ diverges, and if a smaller series diverges any larger series must as well.  Is this right? If I've made any mistakes in any part, please tell me!
29779
I have to test the convergence of the following series:- $$\sum_{n=1}^\infty\sqrt[3]{n^3+1}-n$$ My approach is as follows :- $$n^3+1&gt;1=\sqrt[3]{n^3+1}&gt;1=\sqrt[3]{n^3+1}-n&gt;1-n$$ Now since$\sum 1-n$ diverges, the series under consideration diverges. Is this right or wrong?
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?
19946
If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?  $$\frac {1}{n\log n} \lt \frac {1}{n}$$   $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful. Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?   And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?
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!
55725
I'm struggling to prove that $$\sum \limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}$$ diverges.   Does anyone have any idea how to prove it? Breaking it into smaller pieces din't work. Maybe I should bound it with another series? But how?
69177
Do the following series converge if $a_n&gt;0$ and $\sum_{n=1}^{\infty}a_n$ diverges ?  a.) $\sum_{n=1}^{\infty}{a_n \over 1+ a_n}$  b.)$\sum_{n=1}^{\infty}{a_n\over 1+ a_n ^2}$  c.)$\sum_{n=1}^{\infty}{a_n\over 1+ na_n }$  d.)$\sum_{n=1}^{\infty}{a_n\over 1+ n^2a_n }$  What I thought was using Alembert criteria , but it just not seem to work. I also now if $a_n$ converges so does $a_n^2$. Ive tried everything to no avail, can anyone please help ?
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.)