Question

Consider the series \(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} \)

(a) Find the partial sums \({s_1}\), \({s_2}\), \({s_3}\),and \({s_4}\).Do you recognize the denominators? Use the pattern to guess for \({s_n}\).

(b) Use mathematical induction to prove your guess.

(c) Show that the given infinite series is convergent, and find its sum.

Short answer

• The denominator is\((n + 1)!\)and the numerator is\((n + 1)! - 1\)
• The sequence of partial sums converges to a finite number. So, the series\(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} \)converges and\(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} = 1\).

Step-by-step solution

Partial sums of the series

Let’s start with finding the first few partial sums of the series \(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} \).

We have that \({n^{th}}\)and \({\left( {n - 1} \right)^{th}}\)partial sum \({s_n}\), \({s_{n - 1}}\) satisfy the equation \({s_n} = {s_{n - 1}} + \frac{n}{{(n + 1)!}}\)to simplify computations.

\(\begin{aligned}{s_1} &= \frac{1}{{2!}}\\{s_1} &= \frac{1}{2}\\{s_2} &= \frac{1}{2} + \frac{2}{{3!}}\\{s_2} &= \frac{5}{6}\\{s_3} &= \frac{5}{6} + \frac{3}{{4!}}\\{s_3} &= \frac{{23}}{{24}}\\{s_4} &= \frac{{23}}{{24}} + \frac{4}{{5!}}\\{s_4} &= \frac{{119}}{{120}}\end{aligned}\)

For the first four partial sums, the denominator is\((n + 1)!\)and the numerator is\((n + 1)! - 1\)

And, Reasonable conjecture for the general value of \({s_n}\) is \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\)

Proving the conjecture

Here, we use mathematical induction to prove the conjecture that \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\).

Shown that the value is \({s_1} = \frac{1}{{2!}}\).

For inductive step, assume that \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\), and we have to show that \({s_{n + 1}} = \frac{{(n + 2)! - 1}}{{(n + 2)!}}\)

\({s_{n + 1}} = {a_1} + {a_2} + .... + {a_n} + {a_{n + 1}}\)Where \({a_j} = \frac{j}{{\left( {j + 1} \right)!}}\)

\({s_{n + 1}} = ({a_1} + {a_2} + .... + {a_n}) + {a_{n + 1}}\)

\(\Rightarrow {s_n} + \frac{{n + 1}}{{\left( {\left( {n + 1} \right) + 1} \right)!}}\)

We have the value \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\)

\( = \frac{{(n + 1)! - 1}}{{(n + 1)!}} + \frac{{n + 1}}{{\left( {\left( {n + 1} \right) + 1} \right)!}}\)by the inductive hypothesis that \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\).

\(\begin{aligned} &= \frac{{(n + 2)! - (n + 2)}}{{(n + 2)!}} + \frac{{n + 1}}{{(n + 2)!}}\\ &= \frac{{(n + 2)! - (n + 2) + n + 1}}{{(n + 2)!}}\\ &= \frac{{(n + 2)! - n - 2 + n + 1}}{{(n + 2)!}}\\ &= \frac{{(n + 2)! - 1}}{{(n + 2)!}}\end{aligned}\)

Thus, we proved our claim.

By induction, for all \(n\)we have that the \({n^{th}}\)partial sum \({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\).

Proving that infinite series is convergent

From the above result, we can find the sum of our original series \(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} \).

\({s_n} = \frac{{(n + 1)! - 1}}{{(n + 1)!}}\)is the expression for the partial sums.

By taking the limits to find the eventual behavior of the series,

Since \(\frac{{\left( {\frac{1}{{\left( {n + 1} \right)!}}} \right)}}{{\left( {\frac{1}{{\left( {n + 1} \right)!}}} \right)}} = 1\) for \(n \ne 0\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{1 - \left( {\frac{1}{{(n + 1)!}}} \right)}}{1}\)

The Factorial function grows infinitely large, \((n + 1)! \to \infty \)as \(n \to \infty \).

Then, \(\frac{1}{{(n + 1)!}}\)grows infinitely small, \(\frac{1}{{(n + 1)!}} \to 0\).

Applying these to the above limit,

\(\Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{1 - \left( {\frac{1}{{(n + 1)!}}} \right)}}{1} = 1\)

\(\mathop {\lim }\limits_{n \to \infty } {s_n} = 1\)

The sequence of partial sums converges to a finite number. So, the series\(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} \)converges and\(\sum\limits_{n = 1}^\infty {\frac{n}{{\left( {n + 1} \right)!}}} = 1\).