Analysis of the Series ( sum_{n1}^{infty} frac{1}{sqrt{2n-1}} ): Convergence or Divergence?

Welcome to our detailed exploration of the series ( S 1 frac{1}{sqrt{3}} frac{1}{sqrt{5}} frac{1}{sqrt{7}} cdots ). We will rigorously analyze its convergence or divergence using various mathematical techniques.

Understanding the Series

The series in question is of the form ( a_n frac{1}{sqrt{2n-1}} ) for ( n 1, 2, 3, ldots ). Let's begin our analysis with the divergence test.

Divergence Test

A common method to determine the convergence of a series is the divergence test. According to this test, if the limit of the terms of the series does not approach zero, the series diverges.

Let's calculate the limit of ( a_n ):

[lim_{n to infty} a_n lim_{n to infty} frac{1}{sqrt{2n-1}} 0]

Since the limit is zero, we cannot conclude divergence from this test alone. We need to explore further.

Comparison Test

Next, we can use the comparison test. We compare our series with the series ( sum_{n1}^{infty} frac{1}{sqrt{n}} ). We know that:

[frac{1}{sqrt{2n-1}} sim frac{1}{sqrt{2n}} text{ as } n to infty]

And:

[frac{1}{sqrt{2n}} frac{1}{sqrt{2}} cdot frac{1}{sqrt{n}}]

Since the series ( sum_{n1}^{infty} frac{1}{sqrt{n}} ) is a p-series with ( p frac{1}{2} ), and since ( p leq 1 ), the series diverges.

Hence, because ( frac{1}{sqrt{2n-1}} ) behaves like ( frac{1}{sqrt{n}} ) for large ( n ) and diverges, we can conclude that our original series also diverges.

Infinite Sums and Series Comparison

Let's explore the inequality:

[frac{1}{sqrt{n}} geq frac{1}{n} text{ for } n geq k]

This is equivalent to:

[frac{1}{sqrt{2n-1}} geq frac{1}{n-1} text{ for } n geq k]

Let's sum up left and right sides:

[sum_{i0}^{infty} frac{1}{sqrt{2k 2i}} geq sum_{i0}^{infty} frac{1}{k i}]

By summing the series, the left-hand side (LHS) can be expressed as:

[sum_{n0}^{infty} sqrt{frac{1}{2}n 1}]

And the right-hand side (RHS) is the harmonic series:

[sum_{i0}^{infty} frac{1}{i}]

Since the harmonic series diverges, the original series also diverges.

Pro Tip: Always use an ellipsis (... ) at the end of your finite series to show that it's infinite.

The sum can be expressed as:

[sum_{n1}^{infty} frac{1}{sqrt{2n-1}}]

Direct Comparison Test

For a rigorous approach, let's use the direct comparison test. Consider the series:

[sum_{n1}^{infty} frac{1}{sqrt{2n}}]

This can be simplified as:

[sum_{n1}^{infty} frac{1}{sqrt{2n}} frac{1}{sqrt{2}} cdot zetaleft(frac{1}{2}right)]

Since the zeta function ( zetaleft(frac{1}{2}right) ) diverges for ( p leq 1 ), the series ( sum_{n1}^{infty} frac{1}{sqrt{2n}} ) diverges.

Using the notation ( a_n ) and ( b_n ) where ( a_n frac{1}{sqrt{2n-1}} ) and ( b_n frac{1}{sqrt{n}} ), we have:

[sum_{n1}^{infty} sqrt{frac{1}{2}n} sim sum_{n1}^{infty} frac{1}{sqrt{n}}]

Since ( a_n ) diverges and ( a_n

Thus, the series ( sum_{n1}^{infty} frac{1}{sqrt{2n-1}} ) is divergent.

Conclusion: The series ( 1 frac{1}{sqrt{3}} frac{1}{sqrt{5}} frac{1}{sqrt{7}} cdots ) is divergent.