# Solving Sum of n+1^n/n^(n+1) - Diverges?

• Titans86
In summary, the formula for the "Solving Sum of n+1^n/n^(n+1) - Diverges" series is ∑<sub>n=1</sub>∞ (n+1)^n/n^(n+1). In order for this series to diverge, the limit of the terms must not approach 0. This can be proven by using the limit comparison test or the ratio test. The series cannot converge and it may have real-life applications in mathematical modeling, analysis, and physics.

#### Titans86

Can I use this logic?

## Homework Statement

I'm wondering if I can use this kind of logic to solve:

$$\sum\frac{(n+1)^n}{n^{(n+1)}}$$ Converges or diverges

## The Attempt at a Solution

$$\frac{(n+1)^n}{n^{(n+1)}} \geq \frac{(n)^n}{n^{(n+1)}}$$

And

$$\frac{(n)^n}{n^{(n+1)}} = n$$ , which diverges

Therefor:

$$\sum\frac{(n+1)^n}{n^{(n+1)}}$$ Diverges

Last edited:
Yes. So long as it's true for all n and n > 0.

## 1. What is the formula for the "Solving Sum of n+1^n/n^(n+1) - Diverges" series?

The formula for this series is n=1∞ (n+1)^n/n^(n+1).

## 2. What are the conditions for this series to diverge?

In order for this series to diverge, the limit of the terms must not approach 0. This means that limn→∞ (n+1)^n/n^(n+1) ≠ 0.

## 3. How can we prove that this series diverges?

This series can be proven to diverge by using the limit comparison test or the ratio test. By comparing it to a known divergent series or by finding the limit of the ratio of consecutive terms, we can show that the series diverges.

## 4. Can this series ever converge?

No, this series cannot converge. Since the limit of the terms does not approach 0, the series will always diverge.

## 5. What are some real-life applications of this series?

This series can be used in mathematical modeling and analysis, particularly in the study of infinite series and sequences. It may also have applications in the field of physics, such as in the study of energy and natural phenomena.