My Math Forum limit of a sum

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 23rd, 2014, 07:43 AM #1 Newbie   Joined: Apr 2014 From: USA Posts: 24 Thanks: 1 limit of a sum What's it equal to: $\displaystyle lim_{n\rightarrow \infty }\sum_{1\leqslant k\leq n}\frac{1}{k\ln (n+k))}$
 May 23rd, 2014, 08:34 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,049 Thanks: 680 Math Focus: Physics, mathematical modelling, numerical and computational solutions I get zero.
 May 23rd, 2014, 08:52 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,943 Thanks: 2268 Math Focus: Mainly analysis and algebra Here's my attempt: For fixed n $$\frac{1}{k \ln{(n+k)}}$$ is decreasing, so $$\sum_{k = 1}^{n-1}{\frac{1}{k \ln{(n+k)}}} \gt \int_1^n{\frac{1}{x \ln{(n+x)}}dx} \gt \sum_{k = 2}^{n}{\frac{1}{k \ln{(n+k)}}}$$ And the integral $$\int_1^n{\frac{1}{x \ln{(n+x)}}dx} \gt \int_1^n{\frac{1}{(n + x) \ln{(n+x)}}dx} = \left. \ln{ \left( \ln{( n + x )} \right) } \right|_1^n = \ln{ \left( \ln {2n} \right) } - \ln{ \left( \ln{ \left(n + 1\right)} \right) } = \ln\frac{\ln {2n}}{\ln\left(n+1\right)}$$ And $$\ln\frac{\ln {2n}}{\ln\left(n+1\right)}$$ diverges as $n \to \infty$ so $$\lim_{n \to \infty }\sum_{k=1}^{ n}\frac{1}{k\ln (n+k))} = +\infty$$ Edit: I've just looked: $$\ln\frac{\ln {2n}}{\ln\left(n+1\right)} = \ln{\left(\frac{\ln n}{\ln\left(n+1\right)}+ \frac{\ln 2}{\ln\left(n+1\right)}\right)}$$ converges to zero. So I have some work to do. Thanks from Roli Last edited by v8archie; May 23rd, 2014 at 09:16 AM.

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