Determine if the series converges or diverges. Justify your answer. Sum of 1/(n*(ln n)^2) from n = 2 to infinity.

Determine if the series converges or diverges. Justify your answer. Sum of 1/(n*(ln n)^2) from n = 2 to infinity.

Determining the Convergence or Divergence of the Series: Sum of 1/(n*(ln n)^2) from n = 2 to Infinity

In the realm of mathematical analysis, determining the convergence or divergence of a series is a crucial task. In this article, we will delve into the series: sum of 1/(n*(ln n)^2) from n = 2 to infinity, and explore whether it converges or diverges, providing a detailed justification for our conclusion.

To begin, let's consider the general form of the series: Σ 1/(n*(ln n)^2), where n ranges from 2 to infinity. This type of series is known as a logarithmic series, and its behavior is closely linked to the properties of the natural logarithm function, ln(n).

The key to determining the convergence or divergence of this series lies in the comparison with the well-known harmonic series, Σ 1/n. It is a well-established fact that the harmonic series diverges, meaning that the sum of its terms approaches positive infinity as the number of terms increases.

Now, let's consider the ratio of the terms in our series to the corresponding terms in the harmonic series:

1/(n*(ln n)^2) / (1/n) = 1/(ln n)^2

As n approaches positive infinity, the value of ln(n) also approaches positive infinity. Therefore, the ratio 1/(ln n)^2 approaches 0 as n goes to infinity.

According to the Comparison Test for series, if the ratio of the terms in a series to the corresponding terms in a divergent series (in this case, the harmonic series) approaches 0 as n goes to infinity, then the original series must converge.

Therefore, we can conclude that the series: sum of 1/(n*(ln n)^2) from n = 2 to infinity converges.

The intuition behind this result is that the terms in our series decay faster than the terms in the harmonic series, which is known to diverge. The presence of the logarithmic term, (ln n)^2, in the denominator causes the terms in our series to approach 0 more rapidly than the terms in the harmonic series.

In summary, the series: sum of 1/(n*(ln n)^2) from n = 2 to infinity converges, and this conclusion is justified by the Comparison Test and the behavior of the ratio of the terms in our series to the corresponding terms in the divergent harmonic series.

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