I have the infinite series $1 + \frac{2}{1!} + \frac{3}{2!} + ... $to $ \infty$
My intuition is that $e$ is involved but I don’t know where...
Asked
Active
Viewed 94 times
0
-
1What techniques have you been taught about to find the sum of other infinite series? We are not here to blindly answer your questions but we can help you get gong or assist if you show us where you are stuck – lioness99a Feb 05 '20 at 12:35
-
Wolfram alpha gives result $2e$ – Peter Feb 05 '20 at 12:36
-
Trying to find a pattern comparable to other standard series like $e^x$ or $a^x$ and accordingly proceeding. – user733666 Feb 05 '20 at 12:37
-
https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee Feb 05 '20 at 12:38
-
2Ah, I think I have an idea now : For example $\frac{7}{6!}=\frac{6}{6!}+\frac{1}{6!}=\frac{1}{5!}+\frac{1}{6!}$ – Peter Feb 05 '20 at 12:38
-
https://math.stackexchange.com/questions/2638073/how-to-solve-this-summation-without-taylor – lab bhattacharjee Feb 05 '20 at 12:38
-
Does this answer your question? Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$ – Lê Thành Đạt Feb 06 '20 at 06:25
3 Answers
3
Your sum is:
$$\sum_{n\ge1}^{}\frac{n}{\left(n-1\right)!}$$ Setting $n-1 \mapsto n$ gives:
$$=\sum_{n\ge0}^{}\frac{n+1}{n!}=1+\sum_{n\ge1}^{}\frac{n+1}{n!}=1+\sum_{n\ge1}^{}\frac{1}{n!}+\sum_{n\ge1}^{}\frac{1}{\left(n-1\right)!}=\sum_{n\ge0}^{}\frac{1}{n!}+\sum_{n\ge0}^{}\frac{1}{n!}=\color{red}{2e}$$
1
Use the definition of the exponential: $$ \begin{align} E=&\sum_{n=0}^\infty \frac {n+1}{n!} \\ E=&\sum_{n=0}^\infty \frac {1}{n!}+\sum_{n=1}^\infty \frac {n}{n!} \\ E=&e+\sum_{n=1}^\infty \frac {1}{(n-1)!} \\ E=&e+\sum_{n=0}^\infty \frac {1}{n!} \\ \implies E=&2e \end{align} $$
user577215664
- 40,625
0
Break each term as $\frac{n}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$ now just add these two series of the form $\sum \frac {1}{(n!)}=e$
aud098
- 416