Prove that $\sum \frac{i+1}{i!+1}<3$ 
Prove that $$\sum_{i=1}^{n}\frac{i+1}{i!+1}<3$$

My approach: If $k=1$ we can see that $$S_{1}=\sum_{k=1}^{1}\frac{k+1}{k!+1}=\frac{1+1}{1!+1}=\frac{2}{2}=1<3$$ Now, suppose that the statement is true for some $n\in \mathbb{N}$, it's to say suppose that $S_{n}=\sum_{k=1}^{n}\frac{k+1}{k!+1}<3$ is true, so we need to show that is true for $n+1$, it's to say we need to show that $S_{n+1}=\sum_{k=1}^{n+1}\frac{k+1}{k!+1}<3$ is also true.
Now, we can see that
\begin{eqnarray*}
S_{n+1}&=&\sum_{k=1}^{n+1}\frac{k+1}{k!+1}\\
&=&\sum_{k=1}^{n}\frac{k+1}{k!+1}+\sum_{k=n+1}^{n+1}\frac{k+1}{k!+1}
\end{eqnarray*}
but, how can I continue from here?
 A: As @Arthur points out we cannot directly use induction over the inequality. This is because the summation is an increasing function of $n$ and the upper bound is constant, so if $P(n)<3$ for some constant $n$ then $P(n+1)=P(n)+m<3+m\not<3$ as $m>0$.
We can prove something stronger.
Claim. For all $n\in\Bbb N$ we have $$\sum\limits_{i=1}^n\frac{i+1}{i!+1}<3-\frac1{n^2}.$$
Proof. Let $P(n)$ be the statement in the claim. Evidently $P(1)$ holds so we suppose that $P(k)$ holds for some $k\in\Bbb N$. Then \begin{align}\sum\limits_{i=1}^{k+1}\frac{i+1}{i!+1}=\frac{k+2}{(k+1)!+1}+\sum\limits_{i=1}^k\frac{i+1}{i!+1}<3+\frac{k+2}{(k+1)!+1}-\frac1{k^2}\end{align} so it suffices to show that $$\frac{k+2}{(k+1)!+1}-\frac1{k^2}<-\frac1{(k+1)^2}\impliedby\frac{k+2}{(k+1)!+1}<\frac{2k+1}{k^2(k+1)^2}.$$ Rearranging yields $$(k+1)!>\frac{k^2(k+1)^2(k+2)}{2k+1}-1\impliedby(k-1)!>\frac{(k+1)(k+2)}2$$ which is true for all $k\ge7$ as $(k-1)(k-2)>k+2$ and $k-3>(k+1)/2$. Checking $P(2)$ through $P(6)$ is straightforward and thus $P(n)$ holds for all $n\in\Bbb N$.
A: Alternative proof without induction:
Note that $\frac{i+1}{i!+1}$ looks like $\frac{i}{i!} = \frac{1}{(i-1)!}$ and $\sum_{i\ge1}\frac{1}{(i-1)!}=e < 3$. Unfortunately $\frac{i+1}{i!+1}>\frac{i}{i!} $, but we can make some adjustment:
$$\frac{i+1}{i!+1} < \frac{i+1}{i!}=\frac{i+1}{i} \cdot \frac{1}{(i-1)!}$$
Therefore
$$\sum_{i=1}^{n}\frac{i+1}{i!+1} < \frac{1+1}{1!+1}+\frac{2+1}{2!+1}+\sum_{i=3}^\infty\frac{i+1}{i!+1} < 1+1+\sum_{i=3}^\infty\frac{i+1}{i} \cdot \frac{1}{(i-1)!}\\
< 2+ \frac{4}{3}\sum_{i\ge 3}\frac{1}{(i-1)!}=2+\frac 43\cdot (e-2) \approx 2.958 < 3. \blacksquare
$$
A: Battle plan (aka hint):
1.) First show by induction that $k!>2^{k-1}$ for $k\geq 3$.
2.) Then we have for $k\geq 1$
$$ \frac{k+1}{k!+1} \leq \frac{2k}{k!+1} \leq \frac{2k}{k!} = 2 \frac{1}{(k-1)!}.$$
3.) Now write for $n\geq 3$
$$ \sum_{k=1}^n \frac{k+1}{k!+1} = 1 + 1 +\sum_{k=3}^n \frac{k+1}{k!+1} 
\leq 2+ \sum_{k=3} \frac{1}{(k-1)!}
<2 + \sum_{k=3}^n (1/2)^{k-2}.$$
Where we used 1.) in the last inequality to estimate by a geometric series. Try to conclude.
