I ask this because I noticed the partial sum $\sum_{n=1}^x {1\over \sqrt{n}}$ is very close to $2\sqrt{x}$, so close in fact that it appears their difference approaches a constant value, like $H_x$ and $\ln x$. However, when I put this limit as is into Wolfram, it said the limit diverged.   But, I found a way to transform the limit into an infinite sum, by using the transformation $f(x) = f(0) + \sum_{n=1}^x f(n) - f(n-1)$, an application of telescoping series to partial sums.  Thus, \begin{align*}\lim_{x\to \infty} \left(2\sqrt{x}- \sum_{n=1}^x {1\over \sqrt{n}} \right) &amp;=  \lim_{x\to \infty} \left( 2\sqrt{0} + \sum_{n=1}^x \left( 2\sqrt{n} - 2\sqrt{n-1} \right) - \sum_{n=1}^x {1\over \sqrt{n}} \right) \\ &amp;= \sum_{n=1}^{\infty} \left(2\sqrt{n} - 2\sqrt{n-1} -{1\over \sqrt{n}} \right) \end{align*}  This sum converges according to Wolfram by comparison test, and according to me by the integral test, but what does it converge to? It converges incredibly slowly; my best guess is $\approx 1.458$
96869
How to find $\lim\limits_{n\to\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac1{\sqrt k}\right)$ ?  And generally does the limit of the integral of $f(x)$ minus the sum of $f(x)$ exist? How to prove that and find the limit?
40298
In my homework I have to find limit (by interpreting it as an appropriately chosen function as the limit of integral sums):  $\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} (1)$. $$ $$ Solution:   $\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)}) = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{n^n(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln(\frac{1}{n} n\sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n}))^{1/n} = \lim_\limits{n\to \infty} \frac{1}{n} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})) = \lim_\limits{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} ln(1+\frac{k}{n}) = \lim_\limits{n\to \infty}  \sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})$ $$ $$ Note that function $f(x)=ln(1+x)$ is continuous in $[-1,1]$. Find integral $\int ln(1+x) dx = (x+1) ln(1+x)-x+C$.  Then $\int_{-1}^{1} ln(1+x)dx = ln(4)-2$. $$ $$ Assume that that limit $(1)$ is A. We show that $\forall \epsilon&gt;0$ $\exists N\in \mathbb{N}$: $\forall n\in \mathbb{N}$ $n\geq N \Longrightarrow $ $|\sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})-A|&lt;\epsilon$.  Suppose that $\mathfrak{T}$ is $[-1,1]$ set of subdivisions. Fix. $\epsilon &gt; 0$. That $f$ is integrative $[-1,1]$, then exist $\delta &gt;0$:   $\forall$ $T = T[x_0,...,x_n]\in \mathfrak{T}$ $\forall \xi=(\xi_1,...,\xi_n)\in[x_0,x_1]\times...\times[x_{n-1},x_n]$  $\lambda(T)&lt;\delta \Longrightarrow |\sum_{k=1}^{n} ln(1+\xi_k) (x_k-x_{k-1})-A|&lt;\epsilon$.  Which $N$ to choose? Is this right solution so far?
34558
I have put his on wolfram and obtained answer as follows:  $\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$  And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$  However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?
123075
Let $a_n=\sqrt[n]n-1$. Does $\sum_{n=1}^\infty a_n$ converge?
39848
Does the following series converge or diverge:  $$\sum_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$  Many thanks!
21647
I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$
122884
Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.   This is my idea.  We have the radius of convergence is $R=1$. And $\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.  Then let $f(x)=\sum_{n=1}^{\infty} \dfrac{x^{n+1}}{n(n+1)}$.  We have $f'(x)=\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ and  $f''(x)=\sum_{n=1}^{\infty} x^{n-1}=\dfrac{1-x^n}{1-x}$  Then if I take the primity of $f''(x)$, I get: $f'(x)=\displaystyle\int\limits_{0}^{x}\dfrac{1-t^n}{1-t}\mathrm{d}t=-\ln(1-x)-\displaystyle\int\limits_{0}^{x}\dfrac{t^n}{1-t}\mathrm{d}t$  Then I will get a sum again, and that's not the goal.  What can I do then? Thank you so much.
47290
This is the other direction of  that reads  Let $a_n&gt;0$ (!) for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty a_n&lt;\infty$ implies $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}&lt;\infty$.  I need to prove that &quot;the converse of the last statement is false&quot;. I don't want the answer, but to ask if I can use the following argument to prove the statement. To start, I'm confused with the statement &quot;the converse of the last statement is false&quot;. Is it enough if I prove, if $\sum_n \sqrt{a_n a_{n+1}}$ diverges, then $\sum_n a_n$ must diverge? If this is true, then the proof is easy since I can choose $a_n=1/n$ and prove that $\sum_n a_n$ converges.
36066
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test. $$ \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} $$
17484
Consider the partial sum of the divergent Harmonic series $H_n = \sum\limits_{k = 1}^{n}\frac{1}{k}$. I recently saw a question which required finding out the asymptotic bounds of $H_n$. Now, I could not find any bound closer than $O(n)$. So I looked up the series and found  Wikipedia article which says that $H_n - \ln n$ approaches a constant (the ). This clearly shows that $H_n = \Theta(\log n)$.  However, considering the question was in an beginners undergraduate algorithms course in CS, I was wondering if there is a simpler proof of this? Is there some clever proof requiring elementary mathematics to show $H_n = \Theta(\log n)$?  PS: This is not homework in the sense that it is not my homework, and the actual question does not ask for a proof, but rather asks a lot of functions to be arranged asymptotically.
72165
$$\sum_{n = 1}^\infty \frac{(-1)^n}{p_n}$$ where $p_n$ is the $n$th prime.  I have computed this to 10000 rather than infinity. My results suggest that convergence does happen but it's very slow. But I can't even be sure about the first few digits: $-0.26959$?  I have looked at the "questions that may already have your answer" and some of the "similar questions," but they involve somewhat different formulas.
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.
21194
So my teacher gave a pretty tough (for me) problem in class today: Does the series $$ \sum_{n=1}^{\infty}\frac {\sin{\frac 1 n}} {\sqrt n} $$ converge or diverge?  So far I've thought of trying the Comparison Test and Ratio Test but couldn't really get anywhere. Intuitively I think it should converge, but how do I actually show this?
45441
I believe there are several ways to prove that $\sum\limits^{\infty}_{n=1}\left(1-\frac{1}{\sqrt{n}}\right)^n$ converges. Can you, please, post yours so that we can learn from you?  HERE IS ONE   Let $n\in\Bbb{N}$ be fixed such that $a_n=\left(1-\frac{1}{\sqrt{n}}\right)^n.$ Then, \begin{align} a_n&amp;=\left(1-\frac{1}{\sqrt{n}}\right)^n \\&amp;=\exp\ln\left(1-\frac{1}{\sqrt{n}}\right)^n\\&amp;=\exp \left[n\ln\left(1-\frac{1}{\sqrt{n}}\right)\right] \\&amp;=\exp\left[ -n\sum^{\infty}_{k=1}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;=\exp\left[ -n\left(\frac{1}{\sqrt{n}}+\frac{1}{2n}+\sum^{\infty}_{k=3}\frac{1}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right)\right]\\&amp;=\exp \left[-\sqrt{n}-\frac{1}{2}-\sum^{\infty}_{k=3}\frac{n}{k}\left(\frac{1}{\sqrt{n}}\right)^k\right]\\&amp;\equiv\exp \left(-\sqrt{n}\right)\exp \left(-\frac{1}{2}\right)\end{align} Choose $b_n=\exp \left(-\sqrt{n}\right)$, so that  \begin{align} \dfrac{a_n}{b_n}\to\exp \left(-\frac{1}{2}\right).\end{align} Since $b_n \to 0$, there exists $N$ such that for all $n\geq N,$ \begin{align} \exp \left(-\sqrt{n}\right)&lt;\dfrac{1}{n^2}.\end{align} Hence, \begin{align}\sum^{\infty}_{n=N}b_n= \sum^{\infty}_{n=N}\exp \left(-\sqrt{n}\right)\leq \sum^{\infty}_{n=N}\dfrac{1}{n^2}&lt;\infty,\end{align} and so, $\sum^{\infty}_{n=1}b_n&lt;\infty\implies \sum^{\infty}_{n=1}a_n&lt;\infty$ by Limit comparison test.
80547
Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$  I was advised to proceed with this problem through substitution but that does not seem to help unless I am substituting the wrong parts. If i substitute the $6$, well then i am just stuck with above.   Any ideas on how to proceed. Also, what is the purpose of stating that it is convergent.
91606
I'm supposed to determine whether this sum diverges or converges and if it converges then find its value:  $$ \sum_{n=2}^\infty \frac{1}{n^2-1}. $$  Using the comparison test I eventually showed that this converges.  But I can't figure out how to show what this sum converges to.  The only sums we actually found values for in my notes are geometric series which this clearly isn't.  I saw that I could use partial fraction decomposition to represent the terms as  $$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$ but that just gets me $\infty - \infty$, so this isn't the way to do it.  I'm not sure how to find the value of this sum.  I don't need the full solution but a hint would be appreciated. Thanks. :)
122212
Let $\{a_n\}$ be a sequence of positive real numbers such that   $a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.   Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...  (A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.   Solution attempt:   Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n&gt;0$, the other root is rejected).   We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}&gt;1$, then $\lim a_n \to \infty$.   Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}&gt;0$. (Again, the other root is rejected due to the same reason).   Hence, $(a_n)$ increases monotonically.   Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$  Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}&lt; 1.3$).  So, option $(A)$ is the correct answer.   Is the procedure correct?   I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.   Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?