May 23rd, 2014, 08:43 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, 09:34 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,084 Thanks: 699 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I get zero.

May 23rd, 2014, 09:52 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 Math Focus: Mainly analysis and algebra 
Here's my attempt: For fixed n $$\frac{1}{k \ln{(n+k)}}$$ is decreasing, so $$\sum_{k = 1}^{n1}{\frac{1}{k \ln{(n+k)}}} \gt \int_1^n{\frac{1}{x \ln{(n+x)}}dx} \gt \sum_{k = 2}^{n}{\frac{1}{k \ln{(n+k)}}}$$ 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. Last edited by v8archie; May 23rd, 2014 at 10:16 AM. 

Tags 
infinite, limit, logarithm, product, sum 
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