Post History
#2: Post edited
by
Snoopy
·
2024-08-06T21:31:45Z (5 months ago)
Another way to solve this problem is to exploit the context of the power series, particularly term-by-term differentiation.- Observe that:
- $$
- e^z=\sum_{k=0}^{\infty} \frac{z^k}{k!}, \quad e^z=\left(e^z\right)^{\prime}=\sum_{k=1}^{\infty} k \cdot \frac{z^{k-1}}{k!}, \quad e^z=\left(e^z\right)^{\prime \prime}=\sum_{k=1}^{\infty} k(k-1) \frac{z^{k-2}}{k!} .
- $$
- Setting $z=1$, one gets:
- $$
- e=\sum_{k=0}^{\infty} \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty} k \cdot \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty}\left(k^2-k\right) \frac{1}{k!} .
- $$
- Adding the second and third equalities, one has $\sum_{k=1}^{\infty} \frac{k^2}{k!}=2e$.
- Another way to solve this problem is to exploit the context of [power series](https://en.wikipedia.org/wiki/Power_series), particularly term-by-term differentiation.
- Observe that:
- $$
- e^z=\sum_{k=0}^{\infty} \frac{z^k}{k!}, \quad e^z=\left(e^z\right)^{\prime}=\sum_{k=1}^{\infty} k \cdot \frac{z^{k-1}}{k!}, \quad e^z=\left(e^z\right)^{\prime \prime}=\sum_{k=1}^{\infty} k(k-1) \frac{z^{k-2}}{k!} .
- $$
- Setting $z=1$, one gets:
- $$
- e=\sum_{k=0}^{\infty} \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty} k \cdot \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty}\left(k^2-k\right) \frac{1}{k!} .
- $$
- Adding the second and third equalities, one has $\sum_{k=1}^{\infty} \frac{k^2}{k!}=2e$.
#1: Initial revision
by
Snoopy
·
2024-08-06T21:30:56Z (5 months ago)
Another way to solve this problem is to exploit the context of the power series, particularly term-by-term differentiation.
Observe that:
$$
e^z=\sum_{k=0}^{\infty} \frac{z^k}{k!}, \quad e^z=\left(e^z\right)^{\prime}=\sum_{k=1}^{\infty} k \cdot \frac{z^{k-1}}{k!}, \quad e^z=\left(e^z\right)^{\prime \prime}=\sum_{k=1}^{\infty} k(k-1) \frac{z^{k-2}}{k!} .
$$
Setting $z=1$, one gets:
$$
e=\sum_{k=0}^{\infty} \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty} k \cdot \frac{1}{k!}, \quad e=\sum_{k=1}^{\infty}\left(k^2-k\right) \frac{1}{k!} .
$$
Adding the second and third equalities, one has $\sum_{k=1}^{\infty} \frac{k^2}{k!}=2e$.