How to prove $ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$ using induction?

Question

This is as far as I get. I get stuck here because both sides to not equal each other, but I am not sure what I am doing wrong.

$$ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$

Assume: $ \sum\limits_{k=1}^{c}\frac{k}{(k+1)!} = 1 – 1/(c+1)!$

Prove: $ \sum\limits_{k=1}^{c+1}\frac{k}{(k+1)!} = 1 – 1/((c+1)+1)!$

$$\sum_{k=1}^c \frac{k}{(k+1)!} + \frac{c+1}{((c+1)+1)!} \stackrel{?}{=} 1- \frac{1}{((c+1)+1)!}$$

$$1 – \frac{1}{(c+1)!} + \frac{(c+1)}{((c+1)+1)!} \stackrel{?}{=} 1 – \frac{1}{((c+1)+1)!}$$

Answer 1

Make a common denominator on your left hand side (noting that $(c+1)+1 = c+2$ and that $(c+2)! = (c+2)(c+1)!$):

\begin{eqnarray} 1 - \frac{1}{(c+1)!} + \frac{c+1}{(c+2)!} &=& 1 - \frac{c+2}{(c+1)!(c+2)} + \frac{c+1}{(c+2)!} \\ &=& 1 - \frac{c+2}{(c+2)!} + \frac{c+1}{(c+2)!} \end{eqnarray}

Can you take it from here?

Answer 2

In your last line, on the left side, multiply the middle fraction by $\frac{(c+1)+1}{(c+1)+1}$. This gives it the same denominator as your last fraction on that left side. Add the terms together, and voila.

Answer 3

Here is the key part of the induction step: \begin{align} \sum_{i=1}^{c+1}\frac{i}{(i+1)!} &= \frac{c+1}{(c+2)!}+\color{blue}{\sum_{i=1}^c\frac{i}{(i+1)!}}\tag{by defn. of $\Sigma$}\\[1em] &= \frac{c+1}{(c+2)!}+\color{blue}{\left[1-\frac{1}{(c+1)!}\right]}\tag{by ind. hyp.}\\[1em] &= 1+\frac{c+1}{(c+2)!}-\frac{(c+2)}{(c+2)!}\tag{manipulate}\\[1em] &= 1-\frac{1}{(c+2)!} \end{align} The induction assumption is used in what is highlighted in blue, and the "key" is to realize that $$ \frac{1}{(c+1)!}=\frac{c+2}{c+2}\cdot\frac{1}{(c+1)!}=\frac{(c+2)}{(c+2)!}. $$

Answer 4

Of course, the series can be evaluated without relying on proving, by induction, equality with a given expression.

Simply note that

$$\frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}$$

The ensuing sum is a "telescoping sum," which is easy to evaluate as $1-1/(n+1)!$.

If we want to use induction, we may still use the "telescoping form" and write

$$\sum_{i=1}^{n+1}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)=\sum_{i=1}^{n}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)+1/n!-1/(n+1)!$$

The sum on the right side is, by induction hypothesis, $1-1/n!$, whereupon adding to the second term yields $1-1/(n+1)!$ thus completing the induction proof!