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A geometric series is a type of series where each term after the first is found by multiplying the previous term by a fixed number, called the common ratio. The general form is \( a + ar + ar^2 + ar^3 + \ldots \). A geometric series can either converge or diverge.

• If the absolute value of the common ratio \( r \) is less than 1, the series converges.
• If \( |r| \ge 1 \), the series diverges.

For example, in the series \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \), it is a variation involving an additional term \( \frac{1}{n} \). Despite this added term, the exponential factor \( 2^n \) causes the series to decay rapidly. This means that even though it is not a "pure" geometric series, its behavior closely follows it, ensuring convergence. Whenever you encounter a series where the terms get smaller quickly due to an exponential factor, it's a good hint that it converges similarly to a geometric series.

The harmonic series is a specific type of series expressed as \( \sum_{n=1}^{\infty} \frac{1}{n} \). It's important to note that although it looks simple, the harmonic series diverges.The main reason for its divergence is that the terms decrease very slowly. While it might seem that adding smaller and smaller fractions will eventually result in a finite sum, this is not the case. The terms do not shrink fast enough to yield a convergent series.In the given exercise, the series \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \) has an element similar to the harmonic series, namely \( \frac{1}{n} \). However, due to the multiplication by \( \frac{1}{2^n} \), it decreases faster than the harmonic series. Hence, this additional exponential term allows convergence. When examining series similar to the harmonic series, pay close attention to any factors that might influence the rate of decrease.

The p-series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) where \( p \) is a positive real number. This series is an excellent example for classifying convergence or divergence based on the exponent \( p \).

• If \( p > 1 \), the p-series converges.
• If \( p \leq 1 \), the p-series diverges.

In the solution's context, we encountered the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), where \( p = 2 \). Since \( p \) is greater than 1, the series converges.This simple test is powerful and can be applied broadly within series analysis. When you see a series in this form, quickly identifying the value of \( p \) offers an immediate understanding of the series' behavior. The combination of two sub-series in the original problem illustrates nicely how knowing the properties of each component helps deduce the convergence of the entire series.