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 May 1st, 2012, 05:59 PM #1 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Series convergence $\sum_{n=1}^{\infty} \frac{3}{n^2}$ then after some integration lim (from b to infint) (-3/b+3)=1 so since the integration of 3/n^2 from one to infinity converges then the summation of infinity n=1 3/n^2 would converge?
 May 1st, 2012, 06:45 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,165 Thanks: 472 Math Focus: Calculus/ODEs Re: Series convergence Yes. We have $f(n)=a_n=\frac{3}{n^2}$ Now, since $f\,:\,$1,\infty\)\,\rightarrow\,\mathbb R_+$ we may state: If $\int_{1}\,^{\infty} f(x)\,dx=\lim_{t \to \infty}\int_{1}\,^{t} f(x)\,dx < \infty=$ then the series converges. So, we compute: $\lim_{t \to \infty}\int_{1}\,^{t} 3x^{\small{-2}}\,dx=-3\lim_{t \to \infty}$$\[\frac{1}{x}$_1^{t}$$=-3\lim_{t \to \infty}$$\frac{1}{t}-1$$=3<\infty=$ Hence, by the integral test, the series converges.
 May 1st, 2012, 06:55 PM #3 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: Series convergence YEY!
 May 1st, 2012, 07:13 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,165 Thanks: 472 Math Focus: Calculus/ODEs Re: Series convergence Using the integral test, can you now determine for what values of x the following converges: $\sum_{n=1}^{\infty}\frac{k}{n^x}$ where $k\in\mathbb R$

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