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.
[Math] Sum the series: $1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$
real-analysissequences-and-series
Related Solutions
This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.
Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:
- $\frac12 |BC||BD| \sin\theta = \frac12 \sqrt{n+1} \sqrt{n} \sin\theta$
- $\frac12 |CD||AB| = \frac12 ( \sqrt{n} - \sqrt{n-1} )$
Equate them gives us:
$$\sin \theta = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}$$
On the other hand,
$$\begin{align} \theta &= \measuredangle CBD = \measuredangle CBA - \measuredangle DBA\\ &= \sin^{-1}\frac{|AC|}{|BC|} - \sin^{-1}\frac{|AD|}{|BD|} = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}} \end{align}$$
We obtain
$$\sin^{-1}\left( \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right) = \sin^{-1}\sqrt{\frac{n}{n+1}} - \sin^{-1}\sqrt{\frac{n-1}{n}}$$
and we have turned the original series into a telescoping one. The partial sum of the first $n$ terms of original series becomes the angle $\measuredangle ABC$. When $n \to \infty$, the line $BC$ becomes vertical and this geometrically justify why the limit of the series is $\frac{\pi}{2}$.
$$\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) +\cdots +\left(\frac{1}{n} - \frac{1}{n+1}\right) $$ $$ = 1-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14 -\cdots +\frac{1}{n} - \frac{1}{n+1}$$ $$ = 1-\left(\frac12-\frac12\right)-\left(\frac13-\frac13\right)-\left(\frac14-\frac14\right)-\space \cdots \space - \left(\frac{1}{n}-\frac{1}{n}\right)-\frac{1}{n+1}$$
Notice how each of the terms in parentheses is zero, so we are left with: $$\boxed{\text{Sum of first n terms: }1-\frac{1}{n+1}}$$
If we want the infinite sum we must take the limit as $n \to \infty$ because $n$ is the number of terms in the sequence. So as $n$ becomes arbitrarily large, $\dfrac{1}{n}$ tends towards $0$ so we get that the sequence of finite sums approaches: $$1-0 = \boxed{1}$$
Not all infinite series need to be arithmetic or geometric! This special one is called a telescoping series.
Best Answer
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
$$=\frac{1}{2}((e^3-1)-(e-1))$$