Sum of infinite series $\frac{k(k+1)3^k}{k!}$

Question

Can someone suggest how to find the infinite series sum for

$$\frac{k(k+1)3^k}{k!}$$ where k goes from $1$ to infinity.

I know that $\sum_0\frac{3^k}{k!}=e^3$ but I'm not sure if that helps here.

See https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22/576978#576978 — lab bhattacharjee, Dec 03 '20 at 14:19

score 5 · Answer 1 · answered Dec 03 '20 at 12:56

5

Hint:let $$f(x)=xe^x=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k!}$$ what is $3f''(3)?$

answered Dec 03 '20 at 12:56

Albus Dumbledore

11,860

1

Thanks, I think I'll take it from here. However, could you tell me how to approach these problems in general? How does one "know" which derivative to consider? – PGupta Dec 03 '20 at 13:32
1

@PGupta the $k(k+1)$ told me that i would have to twice differentiate $x^{k+1}$ then i used the formula you gave. – Albus Dumbledore Dec 03 '20 at 14:45
Makes sense, thanks! – PGupta Dec 03 '20 at 15:26
1

@PGupta glad to have helped – Albus Dumbledore Dec 03 '20 at 15:27

Empy2 · Accepted Answer · 2020-12-03T13:51:05.543

1

$$\frac{k(k+1)3^k}{k!}=\frac{k(k-1)3^k}{k!}+\frac{2k3^k}{k!}=\frac{3^k}{(k-2)!}+2\frac{3^k}{(k-1)!}\\ =9\frac{3^{k-2}}{(k-2)!}+6\frac{3^{k-1}}{(k-1)!}$$ Now change variables

edited Dec 03 '20 at 13:51

answered Dec 03 '20 at 13:45

Empy2

50,853

Sum of infinite series $\frac{k(k+1)3^k}{k!}$

2 Answers2