Convergence of $\sum_{n=1}^{\infty} \frac{\frac{1}{2}+(-1)^n}{n}$

Question

Does the following series converge?: $$\sum_{n=1}^{\infty} \frac{\frac{1}{2}+(-1)^n}{n}$$

Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.

Answer 1

Hint

$$S_n= \left(\sum_{k=1}^n \frac{1}{2k} \right)+\left(\sum_{k=1}^n \frac{(-1)^k}{k} \right)$$

The first series diverges and the second converges by Leibnitz test...

Answer 2

Since $$|\frac{1}{2}+(-1)^n|\geq|(-1)^n|-\frac12=\frac12$$ then $$\sum_{n=1}^{\infty} |\frac{\frac{1}{2}+(-1)^n}{n}|>\dfrac12\sum_{n=1}\dfrac1n$$

Answer 3

Let $A(m)=\frac {\frac {1}{2}+(-1)^{2m}}{2m}+$ $\frac {\frac {1}{2}+(-1)^{2m+1}}{2m+1}.$

We have $A(m)=\frac {3/2}{2m}-\frac {1/2}{2m+1}>$ $\frac {3/2}{2m}-$ $\frac {1/2}{2m}=$ $\frac {1}{2m}.$

Therefore $\sum_{n=1}^{2M+1}\left(\frac {\frac {1}{2}+(-1)^n}{n}\right)=$ $-1/2+\sum_{m=1}^M A(m)>$ $-1/2+\sum_{m=1}^M\frac {1}{2m}$

which $\to \infty$ as $M\to \infty.$