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I'm having a horrible time completing this since this is the first time I'm learning mathematical induction.

So far I have this:

Let $P(n)$ be:

$P(n):\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$

For the base case, $n=1$

$LHS = \sum_{i=1}^{0} {\frac{1}{4i^2-1}} = {\frac{1}{4(1^2)-1}} = {\frac{1}{4-1}} = {\frac{1}{3}}$

$RHS = \frac{n}{2n +1} = \frac{1}{2(1) +1} = \frac{1}{2 +1} = \frac{1}{3}$

Therefore $P(1)$ holds.

Assuming $P(k)$, $P(k)$ holds for $n=k$:

$P(k): \sum_{i=1}^{k} {\frac{1}{4i^2-1}} = \frac{k}{2k +1}$

Proving for $P(k+1)$:

$P(k+1):\\ \\\ \\ \sum_{i=1}^{k+1} {\frac{1}{4i^2-1}}$

$= \sum_{i=1}^{k} {\frac{1}{4i^2-1}} + \frac{1}{4(k+1)^2-1}$

$=\frac{k}{2k +1} + \frac{1}{4(k+1)^2-1}$

Can anyone help me from this point onward?

lxkmxl
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    What you've done so far is good. You now need to add the fractions together. It would help to notice that $4(k+1)^2 - 1 = (2k+1)(2k+3)=(2k+1)(2(k+1)+1)$. – Zestylemonzi Apr 14 '17 at 12:18

2 Answers2

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for the sum $$\sum_{i=1}^{k+1}\frac{1}{4i^2-1}$$ we can write: $$\frac{n}{2n+1}+\frac{1}{4(n+1)^2-1}$$ and this must be $$\frac{n+1}{2(n+1)+1}$$ can you proceed?

Dr. Sonnhard Graubner
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Note that the difference between the $(n+1)$th and $n$th putative partial sums is $$\begin{align*}\frac{n+1}{2(n+1)+ 1} - \frac{n}{2n+1} &= \frac{(n+1)(2n+1) - n (2n+3)}{(2n+3)(2n+1)} \\ &= \frac{1}{4n^2 + 8n + 3} \\ &= \frac{1}{4 (n+1)^2 - 1}\end{align*}$$ and this is the $(n+1)$th term in the series.

Connor Harris
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