This is quite infuriating. Just when I feel like I'm grasping this whole series konverging/diverging tests-notion, this one pops up. Here is what I did:  $$\sum_{k=1}^{\infty}|(-1)^k(\sqrt{k+1}-\sqrt{k})|=\sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k})=\sum_{k=1}^{\infty}\frac{1}{\sqrt{k+1}+\sqrt{k}}\leq\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}=\text{Div.}$$  So the series is not absolutely convergent. So this test didin't give me anything.  For Leibniz test, that states that if $\{a_k\}^{\infty}_{k=1}$ is a decreasing sequence of positive numbers with limit equal to zero, then the alternating series $\sum(-1)^{k-1}a_k$ is convergent.  However, $a_k=\sqrt{k+1}-\sqrt{k}$ is unfortunately increasing.
122949
Please suggest some hint to test the convergence of the following series $$\sum_{n=1}^\infty (-1)^n(\sqrt{n+1}-\sqrt n)$$
93437
We know that the "Harmonic Series" $$ \sum \frac{1}{n}$$ diverges. And for $p >1$ we have the result that the series converges $$\sum \frac{1}{n^{p}}$$ converges.   One can then ask the question of testing the convergence the following 2 Series:  $$\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}, \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \sin{n}}}$$ where $ k \in (0,2)$.  Only thing which i have as tool for this problem is the inequality $| \sin{n} | \leq 1$, which i am not sure whether would applicable or not.
89378
Test the series whose general term is: $$\sqrt{n^2 + 1} - n$$   This is from Higher Algebra by Henry Sinclair. I'm struggling to understand how this series is divergent. If you do the following expansion: $$\sqrt{n^2 + 1} - n$$ $$= n\biggl(1 + \frac{1}{n^2}\biggr)^\frac{1}{2} - n$$ $$= n + \frac{1}{2n} - \frac{1}{8n^3} + ... - n$$ $$= \frac{1}{2n} - \frac{1}{8n^3} + ... $$  The remaining series appears to converge to zero as $n$ approaches infinity.  Their explanation is similar to mine, but at the end they say that the above series approaches $\frac{1}{2n}$. From there they directly say that the series is divergent. Earlier in the chapter, it's proved that the expansion of $(1 + x)^n$ is convergent for $x &lt; 1$. In this case, $\frac{1}{n^2} &lt; 1$ as $n$ approaches infinity, so I would assume that the series converges.  The solution to the problem is given  (number 17)  Any help is appreciated. Thanks.
33086
Let $\sum a_n$ be convergent but not absolutely convergent, $\sum_{n=1}^{\infty} a_n=0$,$s_k$ denotes the partial sum then could anyone tell me which of the following is/are correct?  $1.$ $s_k=0$ for infinitely many $k$  $2$. $s_k&gt;0$ and $&lt;0$ for infnitely many $k$  3.$s_k&gt;0$ for all $k$  4.$s_k&gt;0$ for all but finitely many $k$  if we take $a_n=(-1)^n{1\over n}$ then $\sum a_n$ is convergent but not absolutely convergent,but I don't know $\sum_{n=1}^{\infty} a_n=0$? so I am puzzled could any one tell me how to proceed?
34047
Stumped on yet another absolute convergence problem. (Converge conditionally, absolutely, or diverges)  $$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} $$  First, I tested for absolute convergence, and the series reduced to $\frac{\ln(n)}{n}$   .....Using direct comparison, $\frac{\ln(n)}{n} &gt; \frac{1}{n}$ I determined this diverges.  So, it does not converge absolutely.  Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing.  Well, they are not:  $$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} = -0 + .35 - .37 + .35.... $$  So, without using the alternating series test, how do I determine if the orig. series converges or diverges?  How can I use the divergence test (Nth term test)?  I'm not sure what the nth term tends to with that (-1) oscillation.  Just focus on $\frac{\ln(n)}{n}$  But we established that diverges, right?   But, I am told this series converges conditionally.  Do I use integral test?
35907
Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am stuck on this problem. Check the following series for convergence/divergence $$\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$$ I tried to prove the convergence by comparison test  $$\sqrt{n^3+1}-\sqrt{n^3-1}= \frac{2}{\sqrt{n^3+1}+\sqrt{n^3-1}}=\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}$$ and then compare it with $$\sum \limits_{n=1}^{\infty}\frac{1}{n^2}$$  But in order to do that, I need to prove that $$\frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}} \leq 1$$ But I am having problems to prove that. Does anyone have tip how to solve this problem?
24718
$\newcommand{\FUNC}[1]{#1 \frac{nx}{1+n^4x^2}}$  The following problem has bothered me for a very long time. I've been trying to solve this problem for 3 days long - and it feels I tried everything. To conclude - I absolutely need the help of the professionals; And that's why I came here.  I am given with the following function series:  $$S(x)\equiv\sum_{n=1}^{\infty}(-1)^nf_n(x)\equiv\sum_{n=1}^{\infty}\FUNC{(-1)^n}$$  My task is to decide whether the series $S(x)$ uniformly converges on $[0,\infty)$, or not.  Things I tried:  (1) I tried to use Leibniz's Alternating Series Test. I proved that $f_n(x)$ uniformly converges to $0$, but the problem is that $f_n(x)$ is not a decreasing sequence - for every $n\in\mathbb{N}$, there exists $0&lt;x_0\in\mathbb{R}$, such that $f_n(x)$ is increasing on $[0,x_0]$, and decreasing on $[x_0,\infty)$. I've noticed that $\lim_{n\to\infty}x_0=0$, but that doesn't help - since it would never actually reach $0$ (that would only help to prove that the series uniformly converges on $[c,\infty)$ when $c&gt;0$).  (2) I tried to use Weierstrass's M Test. I found that $f_n(x)\leq\frac{1}{2n}$ for every $n\in\mathbb{N}$, but that of course won't help (since $\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges). The huge problem using the M test is that $\frac{1}{2n}$ is actually a tight bound.  (3) I proved that $\displaystyle \sum_{n=1}^{\infty}\FUNC{}$ uniformly converges on $[c,\infty)$ when $c&gt;0$, but does not uniformly converges on $[0,\infty)$. That wasn't very helpful, since $S(x)$ might converges conditionally.  (4) Defining:  $$S_m(x)\equiv\sum_{n=1}^{m}\FUNC{(-1)^n}$$  It is sufficient to show that:  $$(*) \lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|S_m(x)-S(x)\right|\right)\equiv\lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\right)=0$$  in order to prove the series uniformly converges on $[0,\infty).$ Looks impossible and complicated - yet I was eager to try that too. But to no avail.  I will say that my intuition is that the series does uniformly converge on $[0,\infty)$, and that is because I used graphic calculators, and found that for every $n\in\mathbb{N}$:  $$\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\leq\sum_{n=m+1}^{\infty}\frac{1}{n^2}$$  And since $\displaystyle \sum_{n=m+1}^{\infty}\frac{1}{n^2}$ converges, we can say that its limit when $m\to\infty$ is $0$, thus proving $(*)$. However - I couldn't prove this inequality. I tried using the triangle inequality, to distinguish between odd and even indices - nothing worked.  As you might see - I really need help. Thank you very much!
117044
Show that the series converges, but not absolutely. $\sum_{n=1}^{\infty}( $exp$(\frac{(-1)^n}{n})-1)$.   My Try:  Let $a_n=$exp$(\frac{(-1)^n}{n})-1$. I was going to use alternating series test because the sequence $\{a_n\}$ is alternating. But $\{|a_n|\}$ is not decreasing to $0$. So, I am stuck now. Can anybody please give me a hint?
50928
I have come across the following few questions on past exams papers.. I know how to solve these type when it is of the form $a_nz^n$ but don't have a clue what to do with these. Any help would be greatly appreciated!  (i) $\displaystyle\sum_{n=0}^\infty \bigg(\dfrac{1}{2n}\bigg)z^{3n}$.  (ii) $\displaystyle\sum_{n=0}^\infty(2^n+i)(z-i)^n$.  (iii) Show that the radius of convergence of $\displaystyle\sum_{n=0}^\infty z^{n!}$ is $1$ and that there are infinitely many $z \in \mathbb{C}$ with $|z|=1$ for which the series diverges.
46652
Why does the series $\sum_{n=1}^∞ \ln ({n \over n+1})$ diverges? I'm looking for an answer using the comparison test, I'm just not sure what I can compare it to.   And can I have some tips on what to look at when handling with series that have logarithms in the expression?   Thanks in advance!
27086
$$\sum_{n=1}^\infty \frac{i^n}{n}$$  I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
38229
Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?  Thanks!
61698
This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.
34983
$$\sum_{n=1}^\infty {{4^n n!n!}\over{(2n)!}}$$  I tried the ratio test but got that the limit is equal to 1, this tells me nothing of whether the series diverges or converges. if I didn't make any errors when doing the ratio test,  it may diverge, but I need help proving that. Is there any other test I could try.
101242
Written, in Ex.8 Ch.9.1 of the book Advanced Calculus by P. M. Fitzpatrick :     Suppose that $\sum\limits_{k=1}^\infty a_k$ and $\sum\limits_{k=1}^\infty b_k$ are series of positive numbers such that $$\lim_{n \to \infty} \frac{a_n}{b_n}=l \ \ \ \text{and} \ l&gt;0.$$ Prove that the series $\sum\limits_{k=1}^\infty a_k$ converges iff the series $\sum\limits_{k=1}^\infty b_k$ converges.   Am I correct by the following (sketch of) proof? :  1- For a given $\epsilon_1$ there is $N_1$ such that $\left|\frac{a_n}{b_n} - l\right| &lt; \epsilon_1$ for all $n \ge N_1$.  2- Since the series $\sum\limits_{k=1}^\infty a_k$ converges, for a given $\epsilon_2$ there is $N_2$ such that $\left|b_{n+1}+\dots+b_{n+k}\right|&lt; \epsilon_1$ for all $n \ge N_2$ any for all natural numbers $k$.  3- Define $N = \max {\{N_1,N_2}\}$.   4- From Step (1), $a_{n+k} &lt; (\epsilon_1+l) b_{n+k}$ for all $n \ge N$ any for all natural numbers $k$. [Also, $a_i$'s and $b_i$'s are all positive]. Thus $a_{n+1}+\dots+a_{n+k} &lt; (\epsilon_1+l) (b_{n+1}+\dots+b_{n+k})&lt; \epsilon_3$. Then the convergence of the series $\sum\limits_{k=1}^\infty b_k$ implies the convergence of the series $\sum\limits_{k=1}^\infty a_k$.  5- For the reverse implication, we use the fact that $\lim\limits_{n \to \infty}\frac{a_n}{b_n}=l \iff \lim\limits_{n \to \infty}\frac{b_n}{a_n}= \frac{1}{l}= l' &gt;0$ and repeat the process this time for a and b exchanged.   6- The Quotient Property For Sequences hold for a nonzero limit in the denominator, but since the limit in the numerator also is zero so we may use the The Quotient Property For Sequences.    Thanks.
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!
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?