Prove that $\sum_{k=0}^{2n} \frac{1}{\pi^n+\pi^k}={\frac{2n+1}{2\pi^n}}.$

Question

Prove that $$\sum_{k=0}^{2n} \frac{1}{\pi^n+\pi^k}={\frac{2n+1}{2\pi^n}}.$$I tried telescoping taking $$ t_k=\frac{1}{\pi^n+\pi^k}$$ but I am unable get the differencing: $t_k=f_k-f_{k-1}.$ Can some one help me in getting this.

Answer 1

Here you give up telescoping and instead use the symmetry rule $$S_N=\sum_{k=0}^{N} t(k) =\frac{1}{2}\sum_{k=0}^{N} \left (t(k)+t(N-k) \right).$$ Next, if one is lucky $[f(k)+f(N-k)]$ simplifies to zero, constant or a simple summand. Here in this case check that $$t(k)+t(2n-k)= \pi^{-n}.$$ Hence we prove that $$S_{2n}=\frac{2n+1}{2\pi^n}.$$

Answer 2

Hint: $$ \frac1{\pi^n+\pi^{n+j}}+\frac1{\pi^n+\pi^{n-j}}=\frac1{\pi^n} $$