Show that: limn→∞∑k=1n1kln⁡(n)=1

Question

Show that:  limn→∞∑k=1n1kln⁡(n)=1

search633504 · Accepted Answer

By approximating the sum with integrals

\ln (n + 1) = \int_{1}^{n + 1} \frac{1}{x} d x \leq \sum_{k = 1}^{n} \leq 1 + \int_{1}^{n} \frac{1}{x} d x = 1 + \ln n

since 1/k is decreasing for

k \geq 1

. We have that

\frac{\ln (n + 1)}{\ln n} = \frac{\ln n + \ln (1 + \frac{1}{n})}{\ln n} \to 1

as

n \to \infty

and we obtain the result.

chukizosv · Accepted Answer

Starting with  limn→∞∑k=1n1klog⁡(n)  Notice that the numerator represents the Harmonic numbers, so we rewrite this as  limn→∞Hnlog⁡(n)  Now, Hn∼log⁡(n)+γ for big n, where γ≈0.5772156649 is the Euler–Mascheroni constant. Plugging this in, we get  limn→∞log⁡(n)+γlog⁡(n)  1+limn→∞γlog⁡(n)=1

RizerMix · Accepted Answer

One can also use the Stolz–Cesàro theorem. Write&amp;nbsp;∑k=1n1klnn=anbn&amp;nbsp;As&amp;nbsp;an+1-anbn+1-bn=1n+1ln(n+1)-lnn=1(n+1)ln(n+1n)&amp;nbsp;(ln((1+1n)n(1+1n)))-1&amp;nbsp;→(ln(e))-1=1Then&amp;nbsp;limn→∞anbn=1

Show that: \lim_{n\to\infty}\frac{\sum_{k=1}^n\frac{1}{k}}{\ln(n)}=1

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