# Find a simple formula for the partial sum

1. The series $\sum_{n=1}^\infty\frac1{4n^2-1}$ telescopes. Find a simple formula for the $k$th partial sum $S_k$, and use it to determine whether the series converges. If it converges, find its value.

I've mostly tried to intuit my way through. Calculating the partial sums to several degrees by varying $n$, I got $$\frac13+\frac1{15}+\frac1{35}+\frac1{63}+\frac1{99}+\cdots$$ and I can see that it is converging to $\frac12$.

I computed the first few multiplicative differences between the values and saw that $\frac13$ was multiplied by $\frac15$ to get $\frac1{15}$, $\frac1{15}$ by $\frac37$, $\frac1{35}$ by $\frac7{11}$.

I see that both the denominator and the numerator of the multiplier is increasing by two each time. So I see that it is certainly converging. But I cannot quite put the pieces together on the formula. Any help?

• Do you know anything about telescoping? Nov 6, 2015 at 2:53
• @corindo Yes, I spoke to my professor and he told me that they work in such a way that only the first term and the last term of the sequence do not cancel out. Nov 6, 2015 at 2:54

Basically the telescoping technique is like this: You have a sequence $a_1, a_2 , \cdots , a_n \cdots$. You find a sequence $b_1, b_2 , \cdots , b_n \cdots$ such that $a_k = b_k - b_{k+1}$. So summing $a_k$ would be like summing

$$\sum_{k=1}^n a_k = \sum_{k=1}^n( b_k - b_{k+1}) = b_1 - b_2 + b_2 - b_3 + b_3 - b_4\cdots + b_n - b_{n+1}.$$

You can see here that the terms $- b_2 + b_2$, $- b_3+ b_3$ etc. vanish when you sum. So you only have to pick up the first and the last term:

$$\sum_{k=1}^n a_k = \sum_{k=1}^n b_k - b_{k+1} = b_1 - b_{n+1}.$$

So what you need to do here is to find $b_n$ such that $\frac{1}{4n^2 - 1} = b_n - b_{n+1}.$

Hint (using partial fractions) :

$$\frac{1}{4n^2 - 1} = \frac{1/2}{2n-1} - \frac{1/2}{2n+1}$$

Now try writing out the first few terms of the series.

Hint: break $\frac{1}{4n^2-1}$ into $\frac{1}{2n-1}\cdot \frac{1}{2n+1}$, and then apply the idea of telescoping sum
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