Chapter 1: Problem 8

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{n^{5}}{5^{n}}$$

Short Answer

Expert verified

The series converges by the Ratio Test.

Step by step solution

Preliminary Test (Divergence Test)

The preliminary test states that if the limit of the terms of the series does not approach zero, the series diverges. Evaluate the limit of the general term: \[ \frac{n^5}{5^n} \] as $ n \to \infty $.

Compute the Limit

Analyze $ \frac{n^5}{5^n} $. Notice that the exponential function $ 5^n $ grows much faster than the polynomial function $ n^5 $. Use L'Hôpital's Rule if necessary to find: \[ \text{lim}_{n \to \infty} \frac{n^5}{5^n} = 0 \]

Apply Ratio Test

The Ratio Test helps determine the convergence of the series. Evaluate \[ \text{lim}_{n \to \infty} \frac{a_{n+1}}{a_n} = \text{lim}_{n \to \infty} \frac{\frac{(n+1)^5}{5^{n+1}}}{\frac{n^5}{5^n}} \] Simplify it to: \[ \text{lim}_{n \to \infty} \frac{(n+1)^5}{5 \cdot n^5} \] \[ = \frac{1}{5} \text{lim}_{n \to \infty} \frac{(n+1)^5}{n^5} = \frac{1}{5} \]

Conclusion from Ratio Test

Since the limit is $ \frac{1}{5} < 1 $, according to the Ratio Test, the series $ \sum_{n=1}^{} \frac{n^5}{5^n} $ converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence Test

The Divergence Test is often the first step in testing series for convergence.
It states that if the limit of the terms of the series does not approach zero, then the series diverges.
This test is straightforward and can help to quickly identify divergent series.
To apply the Divergence Test, you evaluate the limit of the general term of the series as it approaches infinity.
For example, in the series $\frac{n^5}{5^n}$, you need to find $\text{lim}_{n \to \rightarrow \frac{n^5}{5^n}}$.
If this limit is not zero, the series diverges. If it is zero, further testing is needed to confirm convergence or divergence.

L'Hôpital's Rule

L'Hôpital's Rule helps resolve limits where direct substitution results in an indeterminate form like $\frac{0}{0}$ or $\frac{\text{\tiny{\text{∞}}}\text{\tiny{\text{∞}}}}$.
It states that if the limits $\text{lim}_{x \rightarrow a}\frac{f(x)}{g(x)}$ are indeterminate, you can differentiate the numerator and the denominator to find:$\text{lim}_{x \rightarrow a}\frac{f'(x)}{g'(x)}$.
For example, $\frac{n^5}{5^n}$ forms an indeterminate type when $n \rightarrow \text{∞}$.
By applying L'Hôpital's Rule five times, you simplify the limit iteratively:$\text{lim}_{n \rightarrow \text{∞}}\frac{n^4}{5^n}$, then:\,,, etc.
Finally, you get the limit as 0.

Ratio Test

The Ratio Test is a powerful tool to determine the series convergence.
It involves calculating the limit of the absolute value of consecutive terms ratio $\text{lim}_{n \rightarrow \text{∞}}\frac{a_{n+1}}{a_n}$.
To apply it to $\frac{n^5}{5^n}$, evaluate $\text{lim}_{n \rightarrow \text{∞}}\frac{(n+1)^5}{5^{n+1}} \times \frac{5^n}{n^5}$.
This simplifies to $\text{lim}_{n \rightarrow \text{∞}}\frac{(n+1)^5}{5n^5}$ .
When you divide powers and simplify, you find the ratio is $\frac{1}{5}$.
Since $\frac{1}{5} < 1$, the series converges.

Infinite Series

An infinite series adds up infinitely many terms sequentially.
If the partial sums of these terms approach a finite limit, the series converges.
Otherwise, it diverges.
Understanding the behavior of infinite series is crucial in many areas, including calculus and complex analysis.
Various tests like the Divergence Test, L'Hôpital's Rule, and Ratio Test help in determining the convergence or divergence.
Each test is used in different scenarios based on the series' structure.
Identifying which test to apply is key to solving infinite series problems efficiently.