Calculating the limit of the sum.

Question

I tried calculating the sum:

$$ \lim_{n\rightarrow ∞}\:( \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3}+.......+ \frac{1}{n+n}) $$

using the Sandwich Theorem, however only got that the limit is between $0.5$ and $1$, and was unable to go further with it...? Are there other approaches here?

Answer 1

Hint :

$$\rm L = \int_{0}^1 \frac{1}{1+x} \rm dx$$

Answer 2

$\displaystyle \sum_{k=1}^n \frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$

The limit is $\displaystyle \int_0^1\frac{1}{1+x}dx$

Answer 3

write $$\lim_{n\rightarrow ∞} \left( \frac{1}{1+\frac1n} + \frac{1}{1+\frac2n} + \frac{1}{1+\frac3n}+.......+ \frac{1}{1+\frac nn} \right)\frac1n$$ and use riemann sum.

Answer 4

By Harmonic series

$$\sum_{k=1}^n \frac{1}{n+k}=\sum_{k=1}^{2n} \frac{1}{n}-\sum_{k=1}^n \frac{1}{n}\sim \ln 2n-\ln n= \ln 2$$

Answer 5

Other way. We can show this identity

$$ \sum^{n}_{k=1} \frac{1}{k+n} =\sum^{2n}_{k=1}\frac{(-1)^{k+1}}{k} . $$

So by the $ln$ series we have

$$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}=ln(2)$$

Answer 6

Note that $$ \sum_{m=1}^k\frac{1}{m}\stackrel{\text{def}}=H_{k}=\log k+\gamma+o(1) $$ where $\gamma$ is the Euler Mascheroni Constant. Hence $$ \sum_{m=n+1}^{2n}\frac{1}{m}=H_{2n}-H_{n}=\log(2n)+\gamma-\log(n)-\gamma+o(1)\stackrel{n\to\infty}\rightarrow \log(2) $$ since $$ \log(2n)-\log(n)=\log 2. $$

Answer 7

I thought it might be instructive to present an approach that relies only on straightforward arithmetic along with knowledge of the Taylor series $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$.

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Note that we can write

$$\begin{align} \sum_{k=1}^n \frac{1}{k+n}&=\sum_{k=n+1}^{2n}\frac1k\\\\ &=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac1{2k}\right)-\sum_{k=1}^n\frac1k\\\\ &=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\ &=\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}\tag1 \end{align}$$

Using the Taylor series for $\log(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$, we see that

$$\log(2)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\tag2$$

Finally, letting $n\to \infty$ in $(1)$, we see from $(2)$ that

$$\lim_{n\to \infty}\sum_{k=1}^n \frac1{k+n}=\log(2)$$