Question

If $S_n = \frac{1 \cdot 2}{3!} + \frac{2 \cdot 2^2}{4!} + \frac{3 \cdot 2^3}{5!} + \dots$ upto n terms then the sum of the infinite terms is :

• A. 1
• B. $\frac{2}{3}$
• C. $e$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (A)

1

Answer 2

We are given

$$S = \sum_{n=1}^{\infty} \frac{n\,2^n}{(n+2)!}.$$

Notice that

$$\frac{2^n}{(n+1)!} - \frac{2^{n+1}}{(n+2)!} = \frac{2^n}{(n+2)!}\Big[(n+2)-2\Big] = \frac{n\,2^n}{(n+2)!}.$$

Thus, the $n^{\text{th}}$ term telescopes as:

$$T_n = \frac{2^n}{(n+1)!} - \frac{2^{n+1}}{(n+2)!}.$$

Summing from $n = 1$ to $N$:

$$S_N = \sum_{n=1}^N \left(\frac{2^n}{(n+1)!} - \frac{2^{n+1}}{(n+2)!}\right) = \frac{2^1}{2!} - \frac{2^{N+1}}{(N+2)!}.$$

Taking the limit as $N\to\infty$ (and noting $\frac{2^{N+1}}{(N+2)!}\to 0$):

$$S = \frac{2}{2!} = \frac{2}{2} = 1.$$

Brief Explanation:
Express the $n^{\text{th}}$ term as a telescoping difference, sum up the series, and take the limit to find $S = 1$.