Sum of series 6.

\(\displaystyle \frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\cdots\)

SOLUTION :

We can also write it as

\(\displaystyle \sum_{n=1}^{\infty} \frac{1+2+3+4+\cdots +n}{n!}\)

\(\text{Let}\;\; \displaystyle a_n =\frac{1+2+3+4+\cdots+n}{n!}\)

\(\displaystyle\Rightarrow \quad a_{n}=\displaystyle \frac{\frac{n}{2}[n+1]}{n !}\)

\(\displaystyle\Rightarrow a_{n}=\frac{n(n+1)}{2 n !}\)

\(\displaystyle\Rightarrow a_{n}=\frac{(n+1)}{2(n-1)!}\)

\(\displaystyle\Rightarrow a_{n}=\frac{n}{2(n-1) !}+\frac{1}{2(n-1)!}\)

Add and subtract 1 to numerator

\(\displaystyle\Rightarrow a_{n}=\frac{(n-1)+1}{2(n-1) !}+\frac{1}{2(n-1)!}\)

\(\displaystyle\Rightarrow a_{n}=\frac{1}{2(n-2)!}+\frac{1}{2(n-1) !}+\frac{1}{2(n-1)!}\)

Taking summition on both the sides \(\displaystyle \sum_{n=2}^{\infty}\)

\(\displaystyle\Rightarrow \sum_{n=2}^{\infty} a_{n}=\frac{1}{2} \sum_{n=2}^{\infty} \frac{1}{(n-2) !}+\sum_{n=2}^{\infty} \frac{1}{(n-1) !}\)

Adding \(a_{1}\) On both Sides

\(\displaystyle\Rightarrow a_{1}+\sum_{n=2}^{\infty} a_{n}=\frac{1}{2} \sum_{n=2}^{\infty} \frac{1}{(n-2) !}+\sum_{n=2}^{\infty} \frac{1}{(n-1) !}+a_{1}\;\;\cdots (1)\)

After this we know that \(\;\displaystyle a_{1}+\sum_{n=2}^{\infty} a_{n}=\sum_{n=1}^{\infty} a_{n}\)

And also we have \(a_1=1\) so using all these in (1) We get

\(\displaystyle\Rightarrow \sum_{n=1}^{\infty} a_{n}=\frac{1}{2}[e]+[e-1]+1\)

\(\displaystyle=\frac{1}{2} e+e=\frac{3 e}{2}\)

ANSWER \(\displaystyle \;\; \frac{3e}{2}\)