0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Yoshi May 11 '14 at 3:58
  • $\begingroup$ Just by using the binomial identity $a^2-b^2=(a-b)(a+b)$ twice. $\endgroup$ – LutzL May 11 '14 at 7:31
  • $\begingroup$ 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. $\endgroup$ – Yoshi May 11 '14 at 7:55
  • $\begingroup$ 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. $\endgroup$ – LutzL May 11 '14 at 9:41
  • $\begingroup$ 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. $\endgroup$ – Yoshi May 11 '14 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.