Question

In Exercises \(35-38,\) write an equivalent series with the index of summation beginning at \(n=1\). $$ \sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$

Short answer

The equivalent series of the given series with the index of summation beginning at \(n=1\) is \(\sum_{n=1}^{\infty} \frac{x^n/n}{n !}\).

Step-by-step solution

Write down the given series

First, write down the given series \(\sum_{n=0}^{\infty} \frac{x^{n}}{n !}\). The index of series in this series starts at 0. Our aim is to convert this into a series where index starts at 1.

Substitute \(n\) with \(n-1\)

The trick to write an equivalent series with the index starting from \(n=1\) is to substitute \(n\) with \(n-1\) in the given series. This will not change the value of the series but the index starting point will shift by 1. So, in place of \(n\), we substitute \(n-1\), and the series becomes \(\sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1) !}\).

Simplify the series

Simplify the series by multiplying the numerator and denominator by \(n\), such that \(x^{n-1} = x^n/x\) and \((n-1)! = n!/n\). This gives us the equivalent series as \(\sum_{n=1}^{\infty} \frac{x^n/n}{n !}\).