# Evaluating Limit $$\frac{\ln2}{2}+\cdots+\frac{\ln n}{n}$$

• MHB
• Vali
In summary, the conversation discusses the application of Stoltz-Cesaro to find the limit of a given expression. After applying the method, the limit is found to be $1/2$ instead of 0 as initially thought. The conversation ends with clarification on the approach used to find the limit.
Vali
Hi,

$$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}.$$
After I applied Stoltz-Cesaro I got $$\lim_{n \to \infty}\dfrac{\dfrac{\ln2}{2}+\dfrac{\ln3}{3}+\cdots+\dfrac{\ln n}{n}}{\ln^2 n}=\lim_{n \to \infty}\dfrac{\dfrac{\ln (n+1)}{n+1}}{\ln^2 (n+1)-\ln^2 n}$$
How to continue ? The limit shouldn't be 0 ? because$\lim_{n \to \infty}\frac{ln(n+1)}{n+1}=0$
It's not 0, it's $1/2$ and I don't know why.

I did it, thanks!

Everything is right except the part where the denominator/numerator limits are taken separately (i.e. the very last part).

## What is the purpose of evaluating the limit $\frac{\ln2}{2}+\cdots+\frac{\ln n}{n}$?

The purpose of evaluating this limit is to determine the convergence or divergence of the series. This can help us understand the behavior of the series as the number of terms increases, and can also help us make predictions about the behavior of similar series.

## What is the formula for evaluating this limit?

The formula for evaluating this limit is $\lim_{n\to\infty}\sum_{k=2}^{n}\frac{\ln k}{k}$. This means that we take the limit as the number of terms, n, approaches infinity, and sum the terms from k=2 to n, where ln k is divided by k.

## How do we determine if this series converges or diverges?

We can use the integral test or the comparison test to determine the convergence or divergence of this series. If the integral of the series converges, then the series also converges. Similarly, if the series can be compared to a known convergent or divergent series, we can determine its behavior.

## What is the significance of the natural logarithm in this limit?

The natural logarithm, ln, represents the inverse of the exponential function, e^x. In this limit, it is used to measure the rate of growth of the series. As the number of terms increases, the natural logarithm also increases, indicating the growth rate of the series.

## Can this limit be evaluated for other values besides ln2?

Yes, this limit can be evaluated for any base of the logarithm. However, the convergence or divergence of the series may change depending on the value of the base. For example, if we use a base greater than 1, the series will diverge, while if we use a base between 0 and 1, the series will converge.

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