What is the sum of this series?

[opticaltempest]

I need to find the sum of the following series.

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2^n n!}}}\)

Here is my approach but according to Maple it seems incorrect. Where am I
going wrong?

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2^n n!}}} = 7\left[ {\sum\limits_{n = 0}^\infty {3^n } \cdot \sum\limits_{n = 0}^\infty {\frac{1}{{2^n n!}}} } \right]\)

The first series is a geometric series. Therefore its sum can be easily found.
Hence,

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {3^n } = \frac{1}{{1 - 3}} = - \frac{1}{2}\)

The second series is a bit more difficult. Using the fact that

\(\displaystyle \L
e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}\)

will help us find the sum.

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {\frac{1}{{2^n n!}}} = \sum\limits_{n = 0}^\infty {\frac{{\left( {\frac{1}{2}} \right)^n }}{{n!}}} = e^{\frac{1}{2}}\)

I find the sum to be

\(\displaystyle \L
7\left[ { - \frac{1}{2} \cdot e^{\frac{1}{2}} } \right] \approx - 5.77\)

But according to Maple it evaluates to

Can anyone help me on this problem?

Thanks

 

[galactus]

I need to find the sum of the following series.

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2^n n!}}}\)

Here is my approach but according to Maple it seems incorrect. Where am I
going wrong?

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2^n n!}}} = 7\left[ {\sum\limits_{n = 0}^\infty {3^n } \cdot \sum\limits_{n = 0}^\infty {\frac{1}{{2^n n!}}} } \right]\)

The first series is a geometric series. Therefore its sum can be easily found.
Hence,

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {3^n } = \frac{1}{{1 - 3}} = - \frac{1}{2}\)
Oops, this diverges.

The second series is a bit more difficult. Using the fact that

\(\displaystyle \L
e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}\)

will help us find the sum.

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {\frac{1}{{2^n n!}}} = \sum\limits_{n = 0}^\infty {\frac{{\left( {\frac{1}{2}} \right)^n }}{{n!}}} = e^{\frac{1}{2}}\)

I find the sum to be

\(\displaystyle \L
7\left[ { - \frac{1}{2} \cdot e^{\frac{1}{2}} } \right] \approx - 5.77\)

But according to Maple it evaluates to

Can anyone help me on this problem?

\(\displaystyle \L\\\sum_{0}^{\infty}\frac{1}{n!}=e\)

\(\displaystyle \L\\7(e^{\frac{1}{2}})(e)=31.37182349\)



Thanks

 

[opticaltempest]

I'm not seeing how I can simplify it to

\(\displaystyle \L
7 \cdot e^{\frac{1}{2}} \cdot e\)

If go this way

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2{}^n \cdot n!}}} = 7\left[ {\sum\limits_{n = 0}^\infty {\left( {\frac{3}{2}} \right)^n } \cdot \sum\limits_{n = 0}^\infty {\frac{1}{{n!}}} } \right]\)

Then this series still diverges

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {\left( {\frac{3}{2}} \right)^n }\)

What am I missing?

Thanks

 

[opticaltempest]

I see what I am missing now...

\(\displaystyle \L
\sum\limits_{n = 0}^\infty {7\frac{{3^n }}{{2{}^n \cdot n!}}} = 7 \cdot \sum\limits_{n = 0}^\infty {\frac{{\left( {\frac{3}{2}} \right)^n }}{{n!}}} = 7 \cdot e^{\frac{3}{2}}\)