# Investigating the convergence of a series using the comparison limit test

Actually not sure how to approach this... but I may be missing something:

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.

Note: This looks easy, but as I said... there's something not quite connecting. When writing out the summation, I'm not entirely sure if they're compatible, as one would go from k=1 to n, and the other would go from k=1 to n-1, yet still vary with k... so do I need to use another pronumeral for it? Or can I simply use k-1 instead of k?

## 1 Answer

Yes you can do that. It is called an index shift. Or you can just simply leave out the common terms, using the recursive definition of the sum symbol, $$\sum_{k=1}^n c_k-\sum_{k=1}^{n-1}c_k=\left(\sum_{k=1}^{n-1} c_k+c_n\right)-\sum_{k=1}^{n-1}c_k=c_n,$$ to get \begin{align} a_n=x_n-x_{n-1}&=\frac1{\sqrt n}-2\sqrt{n}+2\sqrt{n-1}=\frac1{\sqrt n}-\frac2{\sqrt{n}+\sqrt{n-1}} \\&=...=-\frac1{\sqrt n(\sqrt{n}+\sqrt{n-1})^2}\sim -\frac1{4n^{3/2}} \end{align}

• Thanks! This helps lots, except, how did you go from the second to your third step after "to get"... where you're changing the surds from normal to fractional. I have a slight idea, but the particular algebra doesn't seem to work for me. – Yoshi May 11 '14 at 3:58
• Just by using the binomial identity $a^2-b^2=(a-b)(a+b)$ twice. – LutzL May 11 '14 at 7:31
• I'm not entirely sure if that's legit... I plotted a the functions $f(n) = \dfrac1{\sqrt{n}}-\dfrac2{\sqrt{n}+\sqrt{n+1}}$ and $g(n) = \dfrac1{\sqrt{n}}+2\sqrt{n}+2\sqrt{n+1}$ against each other on Mathematica, and they didn't have the same plot... so they can't be equivalent expressions. This "finding something to compare $a_n$ to is pretty freaking hard. I've been at it for a while now. – Yoshi May 11 '14 at 7:55
• Because in the second term, you changed the subtraction to an addition. And for some reason, you changed $(n-1)$ to $(n+1)$. Check also your original formulas and how they are reflected in the derived formulas. – LutzL May 11 '14 at 9:41
• Ah, sorry about that, I've edited the correct formulae in now. They were correct in mathematica though, and still were not equivalent. I now know that the series: $g(n) = 2n^{2} - n - 2\sqrt{n^{4} - n^{3}}$ converges to 0.25, however I'm at a loss to show how it converges algebraically/analytically... even without knowing the exact limit. – Yoshi May 11 '14 at 9:44