Video: Convergence or Divergence of a Series with Terms as the Ratio of Terms Raised to Irrational Powers

Determine whether the series ∑_(𝑛 = 1) ^(∞) 𝑛^𝜋/𝑛^𝑒 converges or diverges.

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Video Transcript

Determine whether the series the sum from 𝑛 equals one to ∞ of 𝑛 to the power of 𝜋 divided by 𝑛 to the power of 𝑒 converges or diverges.

The question gives us a series and wants us to determine the convergence or divergence of this series. One thing we can notice about the series given to us in the question is that we can simplify the summand by using an exponent law. In general, we have the 𝑛 to the power of 𝑎 divided by 𝑛 to the power of 𝑏 is equal to 𝑛 to the power of 𝑎 minus 𝑏. So, we can simplify the series to be equal to the sum from 𝑛 equals one to ∞ of 𝑛 to the power of 𝜋 minus 𝑒. In particular, both 𝜋 and 𝑒 are constant, so this is the sum from 𝑛 equals one to ∞ of 𝑛 raised to the power of some constant.

This is now very similar to what we call a 𝑃 series. A 𝑃 series is a sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power 𝑃, where 𝑃 is a constant. And we know that for a 𝑃 series, if 𝑃 is greater than one, then the series is convergent. And if 𝑃 less than or equal to one, then the series is divergent. We can actually change our series to be a 𝑃 series by using another one of our rules of exponents, which says that 𝑛 raised to the power of 𝑐 is equal to one divided by 𝑛 to the power of negative 𝑐. So, this gives us that our series is equal to the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of negative one multiplied by 𝜋 minus 𝑒. And we can simplify this by noticing that negative one multiplied by 𝜋 minus 𝑒 is equal to 𝑒 minus 𝜋.

So, we’ve now shown that we can rewrite our series to be equal to the sum from 𝑛 equals one to ∞ of one divided by 𝑛 to the power of 𝑒 minus 𝜋, which is a 𝑃 series where 𝑃 is equal to 𝑒 minus 𝜋. In particular, we have that 𝑒 minus 𝜋 is approximately equal to negative 0.423. Since negative 0.423 is less than or equal to one, we’ve shown that the series given to us in the question is equal to a 𝑃 series where 𝑃 is less than or equal to one. So, we can conclude that the series given to us in the question must diverge.

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