Prove by induction: $\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}$

Question

Prove $$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}.$$

My problem with this is that it doesn't hold for the base case: $n=1$. This question is from the book "Abstract Algebra" by Charles Pinter (Page 212, exercise 7). When I go about proving it, I can't reach my goal which is $\frac{(n+1)!-1}{(n+1)!}=1-\frac{1}{(n+1)!}$, for $n=n+1$.

Answer 1

It is not the correct formula. It should be:

$$\frac{1}{2!}+\frac{2}{3!}+...+\frac{n}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}$$

Now the base case works. For the inductive step you need to prove:

$$\frac{1}{2!}+\frac{2}{3!}+...+\frac{k+1}{(k+2)!}=\frac{(k+2)!-1}{(k+2)!} \qquad \mathrm{given}$$

$$\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}=\frac{(k+1)!-1}{(k+1)!}$$

If you need a hint, leave a comment.

Also, please, do not write for $n=n+1$. Write $n=k+1$ instead.

Answer 2

Here is the key part of the induction argument (moving from the left-hand side of $n=k+1$ to the right-hand side): \begin{align} \sum_{i=1}^{k+1}\frac{i}{(i+1)!} &= \sum_{i=1}^k\frac{i}{(i+1)!}+\frac{k+1}{(k+2)!}\tag{by defn. of $\Sigma$}\\[1em] &= \frac{(k+1)!-1}{(k+1)!}+\frac{k+1}{(k+2)!}\tag{by ind. hyp.}\\[1em] &= \frac{(k+2)[(k+1)!-1]}{(k+2)!}+\frac{k+1}{(k+2)!}\tag{manipulate}\\[1em] &= \frac{(k+2)(k+1)!-k-2+k-1}{(k+2)!}\tag{expand}\\[1em] &= \frac{(k+2)!-1}{(k+2)!}.\tag{simplify} \end{align} Could you follow all of the steps in how the right-hand side was reaching starting from the left-hand side? Do you see how the inductive hypothesis was used? Etc.

Answer 3

$$\frac{1}{2!}+\frac{2}{3!}+...+\frac{n}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!} $$ Name as $p(n)$
1 step:$$n=1 \to p(1):\frac{1}{2!}=\frac{2!-1}{2!} $$
2nd step :assume $$n=k \to p(k):\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}=\frac{(k+1)!-1}{(k+1)!} $$

3rd step :prove for n=k+1 $$n=k+1 \to p(k+1)=\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}+\frac{k+1}{(k+2)!}=\frac{(k+2)!-1}{(k+2)!}$$ so substitute $\frac{1}{2!}+\frac{2}{3!}+...+\frac{k}{(k+1)!}={\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}$ into p(k+1)

$${\color{Red}{\frac{(k+1)!-1}{(k+1)!} }}+\frac{k+1}{(k+2)!}=\\\frac{(k+1)!-1}{(k+1)!}*\frac{k+2}{k+2}+\frac{k+1}{(k+2)!} =\\\frac{(k+1)!(k+2)-1(k+2)}{(k+2)!}+\frac{k+1}{(k+2)!}=\\ \frac{(k+1)!(k+2)-1(k+2)+(k+1)}{(k+2)!} =\\\frac{(k+1)!(k+2)-1}{(k+2)!} =\\\frac{(k+2)!-1}{(k+2)!} $$ the proof is complete now (by induction)

Answer 4

Here's an alternate proof, with generating functions instead of induction. Note that $$ \sum_{i=1}^n \frac{i}{(i+1)!} $$ is the sum of the first $n$ terms of the series \begin{align*} \sum_{i=1}^\infty \frac{i x^i}{(i+1)!} &= x \sum_{i=1}^\infty \frac{i x^{i-1}}{(i+1)!} \\ &= x \frac{d}{dx} \sum_{i=1}^\infty \frac{x^i}{(i+1)!} \\ &= x \frac{d}{dx} \frac{1}{x} \sum_{i=1}^\infty \frac{x^{i+1}}{(i+1)!} \\ &= x \frac{d}{dx} \frac{1}{x} \left[ e^x - x - 1\right] \\ &= x \frac{d}{dx} \left[ \frac{e^x}{x} - 1 - \frac{1}{x} \right] \\ &= x \left[ \frac{xe^x - e^x}{x^2} + \frac{1}{x^2} \right] \\ &= \frac{xe^x - e^x + 1}{x}. \end{align*} To get the generating function for the sum of the first $n$ terms, we then multiply by $\frac{1}{1 - x}$: \begin{align*} \frac{xe^x - e^x + 1}{x(1-x)} &= \frac{(1 - e^x)(1 - x) + x}{x(1-x)} \\ &= \frac{1}{1-x} - \frac{e^x - 1}{x} \\ \end{align*} The $n$th term of $\frac{1}{1-x}$ is $1$ and the $n$th term of $\frac{e^x - 1}{x}$ is $\frac{1}{(n+1)!}$, so the result is $$ \boxed{1 - \frac{1}{(n+1)!}} $$ as desired.