What is the sum of this series given the closed form? 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.
 A: 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. 
A: 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. 
