Question

Sum the series : -

1- ∑ ( -1/5)^k from ( k = 0 to infinty )

2- ∑ ( 3^k - 5^ k ) / 7^k from ( k=1 to infinity )

3- ∑ ( 1/ k+1 ) from ( k=1 to infinity

Answer 1

$$\begin{array}{l} \frac{\sum_{k=0}^{\infty} (-1/5)^k}{\sum_{k=1}^{\infty} 3^k / 7^k - 5^k / 7^k} \\ \frac{\sum_{k=0}^{\infty} \frac{1}{k+1}}{\sum_{k=0}^{\infty} k + 1} \end{array}$$

Solution. First, note that the third row diverges as harmonic, so the sum is equal to $+\infty$. To sum the first and the second, one might use the formula for sum of geometric progression. The first one $\frac{1}{1 + 1/5} = 5/4$, the second is equal to $\frac{3/7}{1 - 3/7} - \frac{5/7}{1 - 5/7} = 3/4 - 5/2 = -7/4$.

Answer. 1) $5/4$, 2) $-7/4$ 3) diverges.