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I posted this question earlier, but as I don't know if a comment reply or edit will refresh this so people actually see, I'm going to post it again in hopes that someone knows what's going on. Here's the initial question:

Replacing the sequence:

$x_{n}=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-2\sqrt{n},\,\,\,\, n=1,2,....$

By the corresponding series, invesigate it's convergence.

Hint: Take $a_{1}=x_{1}$ and $a_{n}=x_{n}-x_{n-1}$ for $n>1$. Then $x_{n}$ is a sequence of partial sums for $\sum_{k=1}^{\infty}a_{k}$. Use an expicit formula for $a_{n}$ and use the comparison in limit test to show that the series converges.

Solution: Computing $x_n - x_{n-1}$, we get:

$x_{n}-x_{n-1} = \left(\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)$

$a_{n} = \left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)$

$a_{n} = \frac{1}{\sqrt{n}}-2\sqrt{n}+2\sqrt{n-1}$

Now, the comparison test states that:

If $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ for some limit $c$, then either both $a_n$ and $b_n$ converge or diverge. Trick is, I'm really struggling to find a series $b_n$ that lets me show that $a_n$ diverges (i.e. that my fraction has a finite limit, hence $a_n$ diverges).

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marked as duplicate by LutzL, John, Daniel Rust, userNaN, amWhy May 11 at 12:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Doing the Comparison in Limit test against this $a_n$ with a $b_n = n^{\frac{3}{2}}$ gives me a limit to infinity of $n-2n^{2}+2n^{\frac{3}{2}}\sqrt{n-1}$ which I know converges (via Mathematica), but which seems to rest on a hair's edge... I can't modify it in any other way (i.e. squeeze thm) in order to show that it converges. –  Yoshi May 11 at 8:37
    
Duplicate to math.stackexchange.com/q/788771/115115 with answer, as stated in the first sentence. –  LutzL May 11 at 9:43

1 Answer 1

We can find upper and lower bounds for $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ...+\frac{1}{\sqrt{n}}$, using upper and lower Riemann sums for the integral of $f(x) = 1/\sqrt{x}$

$$\int_1^{n+1}\frac{dx}{\sqrt{x}} \leq \sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 1+\int_1^{n}\frac{dx}{\sqrt{x}} . $$

Whence,

$$2\sqrt{n+1}-2 \leq \sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 2\sqrt{n}-1 , $$

and,

$$2(\sqrt{n+1}-\sqrt{n})-2 \leq x_n = \sum_{k=1}^{n}\frac{1}{\sqrt{k}} - 2\sqrt{n}\leq -1. $$

It can be shown that $x_n$ is decreasing and bounded -- hence, convergent -- using

$$x_{n+1} - x_n = \frac{1}{\sqrt {n+1}} -2(\sqrt{n+1}-\sqrt{n})= \frac{1}{\sqrt {n+1}}- \frac{2}{\sqrt{n}+\sqrt {n+1}} < 0$$

Since

$$\lim_{n \rightarrow \infty} (\sqrt{n+1}-\sqrt{n})=0.$$

we find

$$-2 \leq \lim_{n \rightarrow \infty}x_n \leq -1$$

To show that the series with terms $a_n = x_n - x_{n-1}$ converges, we use a limit comparison test with a convergent series. Selecting $b_n = \frac{1}{n^{3/2}}$ we see

$$\lim_{n \rightarrow \infty}\frac{a_n}{b_n} = \lim_{n \rightarrow \infty} n^{3/2}(\frac{1}{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1})=-1/4$$.

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We are asked to use this approach, though. –  Yoshi May 11 at 7:07
    
So you are trying to show the telescoping series for $a_n$ diverges? –  RRL May 11 at 7:19
    
It appears to me that your sequence converges. Therefore look for a limit comparison series leading to a finite limit -- to show the series is convergent. In this case $n^{-3/2)$ should work –  RRL May 11 at 7:57
    
I know the sequence converges, but looking at $\lim_{n\rightarrow\infty}\dfrac{a_{n}}{b_{n}}$ where $a_n=\dfrac{1}{\sqrt{n}}-2\sqrt{n}+2\sqrt{n-1}$ and $b_n=???$, I can't find a convergent $b_n$ where this works. The criteria for this test also states that the limit of the fraction must exist... so really, I'm trying to find a $b_n$ such that it is both convergent, and causes the fraction to be convergent (which I don't know until I try). However, I know all p-series cause this particular fraction to be divergent with this $a_n$. –  Yoshi May 11 at 8:09
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@RRL : I would remove the singularity from the integral in the upper bound and use $1+\int_1^n\frac{dx}{\sqrt{x}}=2\sqrt{n}-1$ instead. –  LutzL May 11 at 10:25

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