$s=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+..$ then,is $ S=6$ or $S=8$?

Question

$$S=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+..$$

a)$S=6$

b)$S=8$

c) Series does not converge

d) None of the above

Answer 1

Note that since $k>k-1$ we have

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

Hence the series $\sum_{k=1}^\infty \frac1{k^2}$ converges and is bounded above by $2$.

The answer is d) None of the above.

In THIS ANSWER, I evaluated the series in closed form using straightforward integration including transformation of coordinates. The result is of course $\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}{6}$.

Answer 2

The answer is d): The sum is $\pi^2/6$ if I recall correctly.