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We have the series$$1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$$How can we find the sum$?$

MY TRY: $n$th term of the series i.e $T_n=\frac{3^0+3^1+3^2+\cdots+3^n}{(n+1)!}$. I don't know how to proceed further. Thank you.

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    Sum the geometric series in the numerator and then split the fraction into two pieces. Each piece should be obviously a multiple of the series for an exponential function. – 2017-01-17

2 Answers 2

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The numerator is a geometric sum that evaluates to,

$$\frac{3^{n+1}-1}{3-1}$$

Hence what we have is,

$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(3^{n+1}-1)}{(n+1)!}$$

$$=\frac{1}{2} \sum_{n=1}^{\infty} \frac{3^n-1}{n!}$$

$$=\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{3^n}{n!}- \sum_{n=1}^{\infty} \frac{1^n}{n!} \right)$$

Recognizing the Taylor series of $e^x$ we have

$$=\frac{1}{2}((e^3-1)-(e-1))$$

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    Wait a sec. The 0th term ≠ 0. How were you able to convert the index from taking values starting with 0, to starting with 1? – 2017-01-17
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    I converted $\sum_{n=0}^{\infty} f(n+1)$ to $\sum_{n=1}^{\infty} f(n)$. They are both notations for $f(1)+f(2)+f(3)+...$ right? @NilabroSaha – 2017-01-17
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    Oh. Yes, that's right. My mistake. – 2017-01-17
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The general term is

$$T_n={1+3+\cdots +3^n\over n!}={3^{n+1}-1\over 2}{1\over (n+1)!}$$

Now we can rewrite the series as

$$S_n={1\over 2}\sum_{n=0}^\infty {3^{n+1}\over (n+1)!}-{1\over 2}\sum_{n=0}^\infty {1\over (n+1!}={1\over 2}\left(e^3-1\right)-{1\over 2}(e-1)={1\over 2}\left(e^3-e\right)$$