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PRASHANT KUMAR

Answered question

2022-05-09

Jazz Frenia

Skilled2023-05-05Added 106 answers

To express the series, we start by writing the general term of the series:

a_{n} = \frac{n}{1 \cdot 3 \cdot 5 \cdot . . . \cdot (2 n + 1)}

We can see that the denominator is a product of odd numbers starting from 1 and increasing by 2. Therefore, we can write the denominator as:

1 \cdot 3 \cdot 5 \cdot . . . \cdot (2 n + 1) = \frac{(2 n + 1)!}{2 \cdot 4 \cdot 6 \cdot . . . \cdot (2 n + 2)}

Substituting this into the general term, we get:

a_{n} = \frac{n}{(2 n + 1)! / (2 \cdot 4 \cdot 6 \cdot . . . \cdot (2 n + 2))}

a_{n} = \frac{2 \cdot 4 \cdot 6 \cdot . . . \cdot (2 n + 2)}{(2 n + 1)!} \cdot n

a_{n} = \frac{2^{n + 1} (n + 1)!}{(2 n + 1)!}

Now, we can express the given series as:

\sum_{n = 1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdot . . . \cdot (2 n + 1)} = \sum_{n = 1}^{\infty} a_{n}

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{2^{n + 1} (n + 1)!}{(2 n + 1)!}

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} \frac{2^{n + 1}}{(2 n + 1)!!}

where

(2 n + 1)!! = 1 \cdot 3 \cdot 5 \cdot . . . \cdot (2 n + 1)

is the double factorial.
We can further simplify the series by writing the general term in terms of a power series:

\frac{2^{n + 1}}{(2 n + 1)!!} = \frac{1}{n + 1} \cdot \frac{(2 \ln 2)^{n + 1}}{(n + 1)!}

Therefore, the series can be written as:

\sum_{n = 1}^{\infty} \frac{2^{n + 1}}{(2 n + 1)!!} = \sum_{n = 1}^{\infty} \frac{1}{n + 1} \cdot \frac{(2 \ln 2)^{n + 1}}{(n + 1)!}

\sum_{n = 1}^{\infty} \frac{2^{n + 1}}{(2 n + 1)!!} = \sum_{n = 2}^{\infty} \frac{(2 \ln 2)^{n}}{n!}

\sum_{n = 1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdot . . . \cdot (2 n + 1)} = \frac{4 \ln 2 - 1}{2}

Thus, we have successfully expressed the given series in a closed form using power series and arrived at the final answer.

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