My Math Forum series divergence

 Real Analysis Real Analysis Math Forum

 May 1st, 2013, 02:02 PM #1 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 series divergence How can I prove that the series $x_n=\left(1+\frac{1}{2}\right)^2-\left(1+\frac{1}{3}\right)^2+\left(1+\frac{1}{4}\r ight)^2-\left(1+\frac{1}{5}\right)^2+...$ diverges?
 May 1st, 2013, 10:31 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: series divergence [color=#000000]$\sum_{k=2}^{\infty}(-1)^{k}\left(1+\frac{1}{k}\right)^{2}$, $a_{n}=\left(1+\frac{1}{n}\right)^{2}$ decreases with $\lim_{n\to\infty}a_{n}=1$, using the "Leibniz test" we deduce that the series diverges.[/color]
 May 3rd, 2013, 03:16 AM #3 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Re: series divergence Good afternoon Zardo, Leibniz test states that if $(b_n)$ is a nonincreasing sequence with $(b_n) \rightarrow 0$, then the series $\sum (-1)^n b_n$ converges. But the theorem don't say anything about the case $(b_n) \rightarrow 0$ is false. In other words, if $(b_n) \rightarrow 0$ is false, the series not necessarilly diverges. The theorem is an implication $A \rightarrow B$, not an equivalence "if and only if". Do you agree?
 May 4th, 2013, 11:01 AM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: series divergence [color=#000000]Yes you are right! Flip of the mind . Here is another idea, suppose that $s_{n}=\sum_{k=2}^{n}\frac{(-1)^{k}(k+1)^2}{k^2}$ and that $\lim_{n\to\infty}s_{n}=s$ then $s_{n+1}-s_{n}=\frac{(-1)^{n+1}(n+2)^2}{(n+1)^2}\Rightarrow \lim_{n\to\infty}s_{n+1}-s_{n}=\lim_{n\to\infty}\frac{(-1)^{n+1}(n+2)^2}{(n+1)^2}\neq 0$ on the other hand $s_{n+1}-s_{n}\to s-s=0$, but this is a contradiction since our hypothesis was that $\sum_{k=2}^{\infty}\frac{(-1)^{k}(k+1)^2}{k^2}<\infty=$ and this means that the series diverges. $\color{red}\rule{200}{2}$ Here is the funny part of this problem, I used mathematica to compute this series and it gave the following result $\sum_{k=2}^{\infty}\frac{(-1)^k(k+1)^2}{k^2}=1.2912386054562555$, but on the other hand $\sum_{k=2}^{\infty}\frac{\cos(k\pi)(k+1)^2}{k^2}$ (which is the same) gave the result "the series does not converge"! Weird huh? This means that when trying to compute series with numerical methods, one should be very careful. [/color]
 May 4th, 2013, 07:34 PM #5 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Re: series divergence Thank you, George. I wish you a nice weekend.

 Tags divergence, series

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post SlamDunk Algebra 11 April 18th, 2012 11:21 AM aaron-math Calculus 5 December 17th, 2011 06:27 PM patient0 Real Analysis 5 December 11th, 2010 06:17 AM HammerTime Real Analysis 1 November 20th, 2008 04:18 AM Jamers328 Calculus 1 January 29th, 2008 01:51 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top