Conjecture-a-formula-for-the-infinite-sum-of-the-series-1-3-1-15-1-35-1-2n-1-2n-1-And-prove-the-formula-by-Induction-

Question Number 118488 by Lordose last updated on 18/Oct/20

\boldsymbol Conjecture \boldsymbol a \boldsymbol formula \boldsymbol for \boldsymbol the \boldsymbol infinite

\boldsymbol sum \boldsymbol of \boldsymbol the \boldsymbol series .

\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \cdot \cdot \cdot \frac{1}{(2 \boldsymbol n - 1) (2 \boldsymbol n + 1)}

\boldsymbol And \boldsymbol prove \boldsymbol the \boldsymbol formula \boldsymbol by \boldsymbol Induction .

Answered by Olaf last updated on 18/Oct/20

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

S_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{(2 k - 1) (2 k + 1)}

S_{n} = \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{2 k - 1} - \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{2 k + 1}

S_{n} = \frac{1}{2} \underset{k = 0}{\sum^{n - 1}} \frac{1}{2 k + 1} - \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{2 k + 1}

S_{n} = \frac{1}{2} (1 + \underset{k = 1}{\sum^{n}} \frac{1}{2 k + 1} - \frac{1}{2 n + 1}) - \frac{1}{2} \underset{k = 1}{\sum^{n}} \frac{1}{2 k + 1}

S_{n} = \frac{1}{2} (1 - \frac{1}{2 n + 1}) = \frac{n}{2 n + 1}

Induction :

for n = 1, S_{1} = \frac{1}{1.3} = \frac{1}{3} = \frac{1}{2 (1) + 1}

\Rightarrow the formula is true for n = 1

Now we suppose the formula is true for n .

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

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

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

S_{n + 1} = \frac{2 n^{2} + 3 n + 1}{(2 n + 1) (2 n + 3)}

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

S_{n + 1} = \frac{n + 1}{2 n + 3}

S_{n} true \Rightarrow S_{n + 1} true

Finally, S_{n} = \frac{n}{2 n + 1}, n ⩾ 1

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