What is the sum of this series given the closed form?

Question

The closed form of a series I am trying to identify is: $$ a_n=\frac{250}{2n -1} $$

How could I get the sum of the series equation from this? I am used to geometric sequences and arithmetic sequences and this one is confusing me.

Answer 1

If you mean that the general term of the series is $a_n = \dfrac{250}{2n - 1}$, then the corresponding series IS the sum of the terms $a_i, \;1\leq a_i$ as $i\to \infty$. That is, the corresponding series is:

$$\sum_{n = 1}^\infty \dfrac{250}{2n-1}$$

This series is divergent; as $n \to \infty$, the cumulative sum blows up (is unbounded).

We can, however, compute a sum given some $n$: $S_n = \sum_{k = 1}^n \dfrac{250}{2n-1}$. $S_n$ is not, however, a series.

Answer 2

Suppose you want to find the partial sum $\displaystyle \sum_{n = 1}^m \dfrac{250}{2n-1}$ and then consider its limit as $m\to \infty$. You could try $$\displaystyle \sum_{n = 1}^m \dfrac{250}{2n-1} = \displaystyle \sum_{n = 1}^{2m} \dfrac{250}{n} - \displaystyle \sum_{n = 1}^{m} \dfrac{250}{2n}=250H_{2m}-125 H_{m}$$

where $H_m=\displaystyle \sum_{n = 1}^m \dfrac{1}{n}\approx \log_e(m)+\gamma+\frac{1}{2m}$ is a harmonic number. So the partial sum will be about $125\log_e(m)+245.4$ which slowly increases without limit as $m$ increases.