# Find the limit $\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n}$

I need help finding the following limit.

$$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n}$$

#### Solutions Collecting From Web of "Find the limit $\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n}$"

When you’re summing over a varying number of terms, there’s a good chance it’s a hidden Riemann sum. Factoring out $\frac{1}{n}$,

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k/n+1}.$$

To see what function it is the Riemann sum of, write out the summation explicitly.

$$\sum_{k=1}^n \frac{1}{k/n+1} = \frac{1}{1/n + 1} + \frac{1}{2/n + 1} + \cdots +\frac{1}{n/n + 1}.$$

It is easy to see that the function is $f(x) = \frac{1}{x}$, $x \in [1,2]$. If this isn’t clear, draw a picture and some rectangles, and see that the heights are obtained from the right Riemann approximation and the base is $\frac{1}{n}$. Then

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k/n+1} = \int_1^2 \frac{1}{x} \, \mathrm{d}x = \ln(2) – \ln(1) = \boxed{\ln(2)}.$$

$$\lim_{n‎\rightarrow‎\infty}\sum_{k=1}^{n}\dfrac{1}{n+k}=\lim_{n‎\rightarrow‎\infty}\sum_{k=1}^{n}\dfrac{\dfrac{1}{k}}{\dfrac{n}{k}+1}=\int_{0}^{1}\dfrac{dx}{x+1}=\ln(x+1)\Big|_{0}^{1}=\ln2-\ln 1=\ln 2.$$

Since $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}-\log n=\gamma,$ $$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k+n}=\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}=\lim_{n\to\infty}\log(2n)-\log n=\log 2.$$

Your sum is equivalent to $$\bbox[10px, border: 2px lightblue solid ]{\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k+n} = \int_1^2 \frac{1}{x} \, \mathrm{d}x = \ln 2.}$$

by considering the sum as being a Riemann sum. Factor the $1/n$ out and expand the series term by term to notice that the Riemann sum is the function $f(x) = \frac{1}{x}$ where $x \in [1,2]$.