What would be the answer of following?

Question

The above question was asked in a test. It would really help if you could explain the methods to solve such questions.

Answer 1

Hint $$\sum_{i=0}^{n} \ln (1+\frac{i}{n})^\frac{1}{n}=\frac{1}{n} \sum_{i=0}^{n} \ln (1+\frac{i}{n})=\frac{1}{n} \sum_{i=1}^{n} \ln (1+\frac{i}{n})$$ is a Riemann sum.

Answer 2

As an alternative we have that

$$\left(1+\frac{i}{n}\right)^\frac{1}{n}=e^{\frac1n \ln \left(1+\frac{i}{n}\right)}=e^{\frac i {n^2}-\frac {i^2} {2n^3}+\frac {i^3} {3n^4}+\ldots}=1+\frac i {n^2}-\frac {i^2} {2n^3}+\frac {i^3} {3n^4}+\ldots$$

$$\ln \left(1+\frac{i}{n}\right)^\frac{1}{n}=\frac i {n^2}-\frac {i^2} {2n^3}+\frac {i^3} {3n^4}+\ldots$$

therefore by Faulhaber's formula and Alternating harmonic series

$$\sum_{i=0}^{n} \ln \left(1+\frac{i}{n}\right)^\frac{1}{n} =\sum_{i=0}^{n}\frac i {n^2}-\frac {i^2} {2n^3}+\frac {i^3} {3n^4}+\ldots =\frac 1 {n^2}\sum_{i=0}^{n}i-\frac {1} {2n^3}\sum_{i=0}^{n}i^2+\frac {1} {3n^4}\sum_{i=0}^{n}i^3+\ldots$$ $$\to \frac12-\frac16+\frac1{12}-\frac1{15}+\frac1{30}+\ldots =\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k(k+1)}=$$ $$=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}-\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k+1}=\ln 2-(1-\ln 2)=2\ln 2-1$$