Introduction

Series analysis is a fundamental aspect of mathematical analysis, frequently used in various fields including physics and engineering. In this article, we delve into the convergence properties of a particular alternating series, specifically:

Alternating Series ( sum_{n1}^{infty} frac{(-1)^n}{n sqrt{ln n^1}} )

To determine if this series converges or diverges, we apply the Leibniz Test (Alternating Series Test). The Leibniz test states that an alternating series of the form ( sum (-1)^n a_n ) converges if:

( a_n geq 0 ) ( a_n ) is a decreasing sequence, i.e., ( a_n geq a_{n 1} ) for all ( n ) sufficiently large. ( lim_{n to infty} a_n 0 )

In our case, we define ( a_n frac{1}{n sqrt{ln n^1}} ).

Step 1: Verify the Limit Condition

First, we need to check the limit condition:

[ lim_{n to infty} a_n lim_{n to infty} frac{1}{n sqrt{ln n^1}} ]

As ( n ) approaches infinity, both ( n ) and ( ln n^1 ) approach infinity. Therefore,

[ sqrt{ln n^1} to infty quad text{and thus} quad n sqrt{ln n^1} to infty ]

This implies,

[ lim_{n to infty} frac{1}{n sqrt{ln n^1}} 0 ]

Step 2: Verify the Decreasing Condition

Next, we need to confirm that ( a_n ) is a decreasing sequence. We examine the ratio of ( a_{n 1} ) to ( a_n ):

[ frac{a_{n 1}}{a_n} frac{n sqrt{ln n^1}}{(n 1) sqrt{ln n^2}} ]

We need to show that this ratio is less than or equal to 1 for sufficiently large ( n ).

As ( n ) increases, ( n 1 ) approaches ( n ), so the term ( frac{n}{n 1} ) approaches 1. Additionally,

[ ln(n 1) leq ln n^2 Rightarrow sqrt{ln(n 1)} leq sqrt{ln n^2} ]

This means the term ( frac{sqrt{ln n^1}}{sqrt{ln n^2}} ) also tends to 1 for sufficiently large ( n ).

Therefore, we conclude that ( a_{n 1} leq a_n ) for sufficiently large ( n ), confirming that ( a_n ) is eventually decreasing.

Conclusion

Since both conditions of the Leibniz test are met:

( lim_{n to infty} a_n 0 ) ( a_n ) is decreasing for sufficiently large ( n )

We can conclude that the series

[ sum_{n1}^{infty} frac{(-1)^n}{n sqrt{ln n^1}} ]

converges.

Additional Insights

The analysis above shows that the series converges conditionally. Interestingly, if we change the signs to all positive, the series diverges.

Consider the series:

[ sum_{n2}^{infty} frac{1}{2 cdot frac{1}{n} sqrt{ln n^1}} ]

This series can be approximated by the integral:

[ int_{2}^{N} frac{1}{2 cdot frac{1}{x} sqrt{ln x^1}} , dx ]

By solving the integral:

[ int_{2}^{N} frac{1}{2 cdot frac{1}{x} sqrt{ln x^1}} , dx int_{2}^{N} sqrt{ln x^1} , dx ]

The integral can be simplified using the substitution ( u sqrt{ln x^1} ), which gives:

[ sqrt{ln N^1} - sqrt{ln 2^1} to infty text{ as } N to infty ]

This indicates that the series diverges when all terms are positive.

Generalization

Further, we explore the more general form of the series:

[ sum_{n2}^{infty} frac{1}{n^a sqrt{ln x^1}} ]

This series converges for ( a > frac{1}{2} ) and diverges for ( a leq frac{1}{2} ).