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  • 1 Post By v8archie
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May 23rd, 2014, 08:43 AM   #1
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limit of a sum

What's it equal to: $\displaystyle
lim_{n\rightarrow \infty }\sum_{1\leqslant k\leq n}\frac{1}{k\ln (n+k))}$
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May 23rd, 2014, 09:34 AM   #2
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I get zero.
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May 23rd, 2014, 09:52 AM   #3
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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 10:16 AM.
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