Question

Determine the positive values of \(x\) for which the following series converges: $$ \sum_{n=1}^{\infty} \frac{x^{n / 2} e^{-n}}{n} $$

Short answer

The series converges for positive values of \(x\) such that \(0 < x < e^2\).

Step-by-step solution

Identify the Series Type

The given series is a sum from n=1 to infinity of the term \(\frac{x^{n/2} e^{-n}}{n}\).

Apply the Ratio Test

To determine the convergence, apply the Ratio Test which states that a series \(\sum a_n\) converges if \(L = \lim_{{n \to \infty}} \left|\frac{a_{n+1}}{a_n}\right| < 1\). For the given series, let \(a_n = \frac{x^{n/2} e^{-n}}{n}\).

Compute the Ratio

Calculate the ratio \(\frac{a_{n+1}}{a_n}\): \[ \frac{a_{n+1}}{a_n} = \frac{\frac{x^{(n+1)/2} e^{-(n+1)}}{n+1}}{\frac{x^{n/2} e^{-n}}{n}} = \frac{n \cdot x^{(n+1)/2} e^{-(n+1)}}{(n+1) \, x^{n/2} e^{-n}}.\] Simplify the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{n \cdot x^{1/2}}{n+1} \cdot e^{-1} = \frac{x^{1/2}}{n+1} \cdot e^{-1} \cdot n. \]

Take the Limit

Take the limit as \(n\) approaches infinity: \[ L = \lim_{{n \to \infty}} \left|\frac{x^{1/2} \cdot n}{n+1} \cdot e^{-1}\right| = x^{1/2} \cdot e^{-1} \lim_{{n \to \infty}} \frac{n}{n+1} = x^{1/2} \cdot e^{-1}.\] Simplify the limit: \[ L = x^{1/2} \cdot e^{-1}. \]

Determine the Convergence Condition

For the Ratio Test, the series converges if \(L < 1\). Set up the inequality: \[ x^{1/2} \cdot e^{-1} < 1. \] Solve for \(x\): \[ x^{1/2} < e. \] Square both sides to get: \[ x < e^2. \]

Key concepts

ratio test

The Ratio Test is a powerful tool in determining the convergence or divergence of a series. This test evaluates the limit of the absolute ratio of consecutive terms of the series. Formally, for a series \(\sum a_n\), we define the ratio L as: \[ L = \lim_{{n \to \infty}} \left|\frac{a_{n+1}}{a_n}\right| \].
If \[ L < 1 \], the series converges.
If \[ L > 1 \], the series diverges.
If \[ L = 1 \], the test is inconclusive.
In our problem, we start with \(a_n = \frac{x^{n/2} e^{-n}}{n}\) and calculate \frac{a_{n+1}}{a_n}\. This step suggests we manipulate these terms algebraically to eventually find the limit as n approaches infinity.

limit calculation

Limit calculation is a foundational concept in calculus and is crucial in evaluating series convergence. For our Ratio Test, once we derive \frac{a_{n+1}}{a_n}\, we need to simplify this expression and determine its limit as \(n \to \infty\).
For our specific series: \[ \frac{a_{n+1}}{a_n} = \frac{x^{(n+1)/2} e^{-(n+1)} \frac{n}{(n+1) x^{n/2} e^{-n}}} \],
simplifying further, we obtain: \[ = \frac{n x^{1/2} e^{-1} n}{n+1} \].
By isolating \(n\), this reduces to \[ L = x^{1/2} \cdot e^{-1} \cdot \lim_{{n \to \infty}} \frac{n}{n+1} \].
As \(n\to\infty\), \[ \frac{n}{n+1} \rightarrow 1 \], resulting in \[ L = x^{1/2} e^{-1} \]. Essential in limit calculations is acknowledging behavior as n becomes very large.

inequalities

Understanding inequalities is vital when determining a range of values for which a series converges. For the Ratio Test result, \(L = x^{1/2} e^{-1}\), we seek \(L < 1\) for the series to converge.
This gives us the inequality: \[ x^{1/2} e^{-1} < 1 \].
Solving for \(x\) involves isolating \(x\).
First, multiply both sides by \(e\). This removes the exponential: \[ x^{1/2} < e \].
Finally, square both sides to solve for \(x\): \[ x < e^2 \].
Through these steps, we've used properties of inequalities to find that the series converges for \[ x < e^2 \]. Understanding inequalities assists in evaluating the permissible values for variables in convergence problems.