Ramanujan's 1/pi Series - Proving an inequality relating to the error of partial sums

I am a bit stuck on a question I have regarding Ramanujan's formula:

$$\frac{1}{\pi} = \sum_{n=0}^{\infty} \frac{\sqrt{8}(4n)!(1103+26390n)}{9801(n!)^4396^{4n}}$$

Firstly using the ratio test I was able to show that (for $n \geq1$).

$$a_{n+1} < La_n$$

implies (for $n \geq 2$)

$$a_n< L^{n-1}a_1$$

I just bring this up because this was the previous part of the question. The next part is stated below:

Let $S_n$ be the $n$th partial sum of our series starting with the 0th partial sum $S_0 = a_0$. Show that (for $n\geq 1$):

$$0 < \frac{1}{\pi} - S_n < \frac{a_1L^n}{1-L}$$

This is where I am stuck. From what I have read and tried to understand, this stuff is related to the proof of the ratio test and thus related to the geometric series. I don't really know how to use any of that information though when showing this.

Any help would be appreciated.

• I don't know why a very deep and complicated (but very beautiful at the same time) series from Ramanujan's work was chosen for this question. The result is question is derived just from the inequality $a_{n + 1} < L a_{n}$. Perhaps this was done to intimidate the person who is solving it. – Paramanand Singh May 14 '16 at 16:48
• We have $1/\pi - S_{n} = a_{n + 1} + a_{n + 2} + \cdots$ which is less than $a_{1}(L^{n} + L^{n + 1} + \cdots$ and this can be summed as infinite GP and is equal to $a_{1}L^{n}/(1 - L)$. – Paramanand Singh May 14 '16 at 16:50
• Thank you for your help. Your result makes sense, but I am unsure how you were able to get $\frac{1}{\pi} - S_n = a_{n+1}+a_{n+2}....$. – Floatzel98 May 14 '16 at 22:58
• Your original series is written as $$1/\pi = a_{0} + a_{1} + \cdots + a_{n} + a_{n + 1} + \cdots$$ and by definition of partial sum $$S_{n} = a_{0} + a_{1} + a_{2} + \cdots + a_{n}$$ and hence $$\frac{1}{\pi} = S_{n} + a_{n + 1} + a_{n + 2} + \cdots$$ – Paramanand Singh May 15 '16 at 4:28