Series - The Basics — Question 8

Question 8 – Solution

Step 1: Reindex carefully.

Let $k=n+1$. When $n=1$, the new index is $k=2$, and as $n\to\infty$, so does $k$. Therefore $$\sum_{n=1}^{\infty}\frac1{(n+1)!}=\sum_{k=2}^{\infty}\frac1{k!}.$$

Step 2: Compare with the exponential series.

The Maclaurin series for $e=e^1$ is $$e=\sum_{k=0}^{\infty}\frac1{k!}=\frac1{0!}+\frac1{1!}+\sum_{k=2}^{\infty}\frac1{k!}.$$ The desired tail begins at $k=2$, so the omitted terms are exactly $1/0!=1$ and $1/1!=1$. Therefore $$S=e-2.$$

Step 3: Bound the remainder.

After the original terms through $n=N$, the tail starts at $k=N+2$. For $j\ge0$, each new factorial factor is at least $N+3$, so $$(N+2+j)!\ge (N+2)!(N+3)^j.$$ Taking reciprocals and summing gives $$0<R_N\le\frac1{(N+2)!}\sum_{j=0}^{\infty}\frac1{(N+3)^j} =\frac{N+3}{(N+2)(N+2)!}.$$ This bound is explicit, positive, and tends to zero, confirming the convergence of the partial sums to $e-2$.