# Video: Writing an Exponential Decay Model for a Given Situation

Consider the series ∑_(𝑛 =1)^(∞) sin 𝑛/𝑛³. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

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### Video Transcript

Consider the series the sum from 𝑛 equals one to ∞ of sin of 𝑛 over 𝑛 cubed. Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Recall that a series 𝑎 𝑛 is absolutely convergent if the series of absolute values is convergent. And if we find that the series is not absolutely convergent, it may still be conditionally convergent. So we then test the series for convergence or divergence. So let’s begin by testing this series for absolute convergence. So we want to find out whether the sum from 𝑛 equals one to ∞ of the absolute value of sin of 𝑛 over 𝑛 cubed is convergent or divergent.

Well, because 𝑛 only runs through positive values from one to ∞, 𝑛 cubed is always going to be positive. So this is just the sum from 𝑛 equals one to ∞ of the absolute value of sin of 𝑛 over 𝑛 cubed. Now we know that sin of 𝑛 will always be between negative one and one. So we can say that the absolute value of sin of 𝑛 will always be less than or equal to one, which means that we can write the absolute value of sin of 𝑛 over 𝑛 cubed is less than or equal to one over 𝑛 cubed. Writing it this way allows us to do a direct comparison. Recall that this means if 𝑎 𝑛 is less than 𝑏 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, then the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 also converges.

And one over 𝑛 cubed is actually a series we recognize. Recall that a 𝑃-series is a series of the form the sum for 𝑛 equals one to ∞ of one over 𝑛 to the 𝑃 power. And this converges if 𝑃 is greater than one and diverges if 𝑃 is less than or equal to one. So one over 𝑛 cubed is a 𝑃-series with 𝑃 equal to three. So one over 𝑛 cubed converges. So by direct comparison, the absolute value of sin of 𝑛 over 𝑛 cubed also converges. Then, because we found the series of absolute values to be convergent, then our series the sum from 𝑛 equals one to ∞ of sin of 𝑛 over 𝑛 cubed is absolutely convergent.