Exercises on Numerical Series

A selection of worked examples involving infinite series:

exercise	$ \sum_{n=2}^{\infty} \frac{n+1}{n-1} $
exercise	$ \sum_{n=1}^{\infty} \frac{n}{2n+1} $
exercise	$ \sum_{n=1}^{\infty} \frac{2^n - 1}{2^n} $
exercise	$ \sum_{n=1}^\infty \frac{2n + 1}{n} $
exercise	$ \sum_{k=1}^{+\infty} \sqrt[k]{3} $
exercise	$ \sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)} $

 

Step-by-step solution: Analysis of the series $ \sum_{n=2}^{\infty} \frac{n+1}{n-1} $

In this exercise, I’m analyzing the series

\[ \sum_{n=2}^{\infty} \frac{n+1}{n-1} \]

The index starts at 2 to avoid division by zero.

By inspecting the general term, it becomes clear that it can be rewritten in a more manageable form to facilitate the analysis.

I express the numerator $ n+1 $ as $ n - 1 + 2 $, which yields:

$$ \frac{n+1}{n-1} = \frac{n - 1 + 2}{n - 1} $$

$$ \frac{n+1}{n-1} = \frac{(n - 1) + 2}{n - 1} $$

$$ \frac{n+1}{n-1} = \frac{n-1}{n-1} + \frac{2}{n - 1} $$

$$ \frac{n+1}{n-1} = 1 + \frac{2}{n - 1} $$

This allows us to rewrite the original series as:

$$ \sum_{n=2}^{\infty} \left(1 + \frac{2}{n - 1}\right) $$

Now, applying the linearity of summation, we have:

$$ \sum_{n=2}^{\infty} 1 + \sum_{n=2}^{\infty} \frac{2}{n - 1} $$

$$ = \sum_{n=2}^{\infty} 1 + 2 \cdot \sum_{n=2}^{\infty} \frac{1}{n - 1} $$

This representation is more convenient for determining the convergence of the series.

The first sum, \( \sum_{n=2}^{\infty} 1 \), is the infinite sum of the constant 1, which clearly diverges. The second sum, \( \sum_{n=2}^{\infty} \frac{1}{n - 1} \), is the harmonic series shifted by one index (with $ k = n - 1 $), starting at 2 - and it, too, diverges.

Therefore, we conclude that the series diverges.

$$ \sum_{n=2}^{\infty} \frac{n+1}{n-1} = \underbrace{ \sum_{n=2}^{\infty} 1}_{\text{divergent}} + 2 \cdot \underbrace{ \sum_{n=2}^{\infty} \frac{1}{n - 1} }_{\text{divergent}} $$

Hence, the series $ \sum_{n=2}^{\infty} \frac{n+1}{n-1} $ diverges.

Exploring the convergence of the series $ \frac{n}{2n+1} $ step-by-step

In this exercise, the goal is to determine the behavior of the series

\[ \sum_{n=1}^{\infty} \frac{n}{2n+1} \]

Determining the behavior of a series means establishing whether it converges - that is, whether it approaches a finite sum - or diverges, meaning the sum is either infinite or undefined.

The first step is to analyze the general term of the series:

\[ a_n = \frac{n}{2n+1} \]

To assess convergence, we need to examine whether the general term tends to zero as \( n \to \infty \). This is a necessary condition for convergence.

So we compute the limit of the general term:

\[ \lim_{n \to \infty} \frac{n}{2n+1} \]

According to the necessary condition for convergence, if the series \( \sum a_n \) converges, then it must be that \( \lim_{n \to \infty} a_n = 0 \). If this limit is not zero, the series diverges automatically.

To evaluate the limit, we divide both the numerator and denominator by \( n \) to simplify the expression:

\[ \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2 + 0} = \frac{1}{2} \]

The limit of the general term is not zero, but rather \( \frac{1}{2} \):

\[ \lim_{n \to \infty} a_n = \frac{1}{2} \neq 0 \]

This implies that although the terms are positive, they do not approach zero.

Therefore, the series $ \sum_{n=1}^{\infty} \frac{n}{2n+1} $ is divergent.

Exercise: Analysis of the Numerical Series $ \sum_{n=1}^{\infty} \frac{2^n - 1}{2^n} $

We are tasked with analyzing the behavior of the series:

$$ \sum_{n=1}^{\infty} \frac{2^n - 1}{2^n} $$

The general term of the series is given by:

$$ a_n = \frac{2^n - 1}{2^n} $$

This expression can be simplified as follows:

$$ a_n = \frac{2^n - 1}{2^n} = \frac{2^n}{2^n} - \frac{1}{2^n} = 1 - \frac{1}{2^n} $$

Let’s now evaluate the limit of the sequence \( a_n \) as \( n \) approaches infinity:

$$ \lim_{n \to \infty} a_n $$

$$ \lim_{n \to \infty} \left(1 - \frac{1}{2^n}\right) $$

$$ \lim_{n \to \infty} 1 - \lim_{n \to \infty} \frac{1}{2^n} = 1 - 0 = 1 $$

Since the limit of the general term is not zero, the necessary condition for convergence fails. As a result, the series diverges.

Indeed, we can rewrite the original series as the difference of two known series:

$$ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \left(1 - \frac{1}{2^n}\right) = \sum_{n=1}^{\infty} 1 - \sum_{n=1}^{\infty} \frac{1}{2^n} $$

The first series on the right-hand side clearly diverges (being a sum of infinitely many ones), while the second is a convergent geometric series. The divergence of the first term dominates, hence:

The series is divergent.

Analyzing the Convergence of the Series \( \sum_{n=1}^\infty \frac{2n + 1}{n} \)

We are asked to determine whether the series

\[ \sum_{n=1}^\infty \frac{2n + 1}{n} \]

converges or diverges.

Let’s begin by examining the general term of the series:

\[ a_n = \frac{2n + 1}{n} \]

We can simplify the expression as follows:

\[ a_n = \frac{2n}{n} + \frac{1}{n} = 2 + \frac{1}{n} \]

So each term of the series takes the form:

\[ a_n = 2 + \frac{1}{n} \]

To assess convergence, we consider the limit of the general term as \( n \to \infty \). A necessary condition for the convergence of an infinite series \( \sum a_n \) is that its general term tends to zero:

\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(2 + \frac{1}{n} \right) = 2 \]

Since the general term does not approach zero, but instead tends to 2, this condition is not satisfied:

\[ \lim_{n \to \infty} a_n = 2 \ne 0 \]

Therefore, the series diverges by the divergence (or nth-term) test.

And so on.