<s>[INST] 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?
 [/INST] 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
$$\sum_{i=1}^{n}\frac{i+1}{i!+1} < \frac{1+1}{1!+1}+\frac{2+1}{2!+1}+\sum_{i=3}^\infty\frac{i+1}{i!+1} < 1+1+\sum_{i=3}^\infty\frac{i+1}{i} \cdot \frac{1}{(i-1)!}\\
< 2+ \frac{4}{3}\sum_{i\ge 3}\frac{1}{(i-1)!}=2+\frac 43\cdot (e-2) \approx 2.958 < 3. \blacksquare
$$
</s>