Make the summation element of $\sum_{n=1}^{\infty} \frac{1}{n \cdot e}$ to be $ \frac{1}{n}$?

Question

If I want the summation element of $$\sum_{n=1}^{\infty} \frac{1}{n \cdot e}$$ to be $ \frac{1}{n}$ and completely eliminate $e$, is the following solution correct?

$$\sum_{n=1}^{\infty} \frac{1}{n \cdot e}$$ $$\sum_{\frac{n}{e}=\frac{1}{e}}^{\infty} \frac{1}{\frac{n}{e} \cdot e}$$ $$\sum_{n=1}^{\infty} \frac{1}{n}$$

Answer 1

No. That series diverges, since it is the harmonic series times a non-zero constant (and also because the harmonic series diverges).

Answer 2

Elaborating on zwim 's comment, one has, for all positive integer $k$, $$ \sum_{n = 1}^k \frac{1}{n \cdot e} = \frac{1}{e} \sum_{n=1}^k \frac{1}{n} $$ When $k$ tends to $\infty$, the right hand side diverges, and hence so is the left hand side.