The given series is a special type of series known as a p-series. A p-series is a series of the form ∑(1/n^p), where p is a constant.

The p-series test for convergence states that a p-series converges if p 1 and diverges if p ≤ 1.

In the given series, the terms are of the form 1/(1.3.5...), which can be rewritten as 1/n^p, where n is the term number and p is the product of the first n odd numbers.

As n increases, p also increases, so the terms of the series are getting smaller.

Therefore, the given series is a convergent p-series.

Question

The given series is a special type of series known as a p-series. A p-series is a series of the form ∑(1/n^p), where p is a constant.

The p-series test for convergence states that a p-series converges if p > 1 and diverges if p ≤ 1.

In the given series, the terms are of the form 1/(1.3.5...), which can be rewritten as 1/n^p, where n is the term number and p is the product of the first n odd numbers.

As n increases, p also increases, so the terms of the series are getting smaller.

Therefore, the given series is a convergent p-series.

Knowee AI · Accepted Answer

The given series is a special type of series known as a p-series. A p-series is a series of the form ∑(1/n^p), where p is a constant.

The p-series test for convergence states that a p-series converges if p > 1 and diverges if p ≤ 1.

In the given series, the terms are of the form 1/(1.3.5...), which can be rewritten as 1/n^p, where n is the term number and p is the product of the first n odd numbers.

As n increases, p also increases, so the terms of the series are getting smaller.

Therefore, the given series is a convergent p-series.

The Series 1+ 1/1.3+1/1.3.5+ ……… is convergent

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