# Video: Using Patterns to Determine Convergence of Series

Consider the series ∑_(𝑛 = 1) ^(∞) ln((𝑛 + 1)/𝑛). Give an exact expression for the partial sum ∑_(𝑛 = 1) ^𝑀 ln((𝑛 + 1)/𝑛). Is the series convergent?

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### Video Transcript

Consider the series, the sum from 𝑛 equals one to infinity of the natural log of 𝑛 plus one all divided by 𝑛. Give an exact expression for the partial sum from 𝑛 equals one to 𝑀 of the natural log of 𝑛 plus one all divided by 𝑛. Is this series convergent?

The question is asking us for two things. First, it wants us to give an exact expression for the partial sum, the sum from 𝑛 equals one to 𝑀 of the natural log of 𝑛 plus one all divided by 𝑛. And then, it wants us to consider the convergence of the series given to us in the question, the sum from 𝑛 equals one to infinity of the natural log of 𝑛 plus one divided by 𝑛.

Let’s start by trying to find an exact expression for the partial sum given to us in the question. We will denote this by 𝑆 𝑀. So, we have 𝑆 𝑀 is equal to the sum from 𝑛 equals one to 𝑀 of the natural log of 𝑛 plus one divided by 𝑛. We recall that for a finite sum, we can just add each term in the sum together.

To find the first term, we evaluate the natural logarithm of 𝑛 plus one over 𝑛 at one, giving us the natural log of one plus one divided by one. We then do the same to find the second term, which is the natural log of two plus one all divided by two. We can then get the third term as the natural log of three plus one all divided by three. And then, we continue this summing process up until the end point of our summation, which is when 𝑛 is equal to 𝑀, giving us the final term of the natural log of 𝑀 plus one all divided by 𝑀. We can then evaluate the expressions inside our natural logarithm functions.

Now, we want to use one of our rules for logarithms, which says that the log of 𝑎 plus the log of 𝑏 is equal to the log of 𝑎 multiplied by 𝑏. This means, instead of adding each term of the logarithms as we have before, we can now take the log of the product. Which gives us the natural log of two over one multiplied by three over two multiplied by four over three multiplied by five over four all the way up to being multiplied by 𝑀 plus one divided by 𝑀. Multiplying out our expression gives us a numerator of two multiplied by three multiplied by four multiplied by five all the way up to being multiplied by 𝑀 plus one.

Similarly, if we multiply the denominators, we get one multiplied by two multiplied by three multiplied by four being multiplied all the way up to 𝑀. We can then start canceling the shared factors. We see there’s a shared factor of two and a shared factor of three.

In fact, we notice we can get this pattern of our shared factors across the entire fraction, where the only factor to not get canceled in the numerator is 𝑀 plus one and the only factor to not get canceled in the denominator is one. Giving us the natural logarithm of 𝑀 plus one all divided by one, which is just equal to the natural logarithm of 𝑀 plus one. So, what we have shown is that our partial sum from 𝑛 equals one to 𝑀 of the natural logarithm of 𝑛 plus one all divided by 𝑛 is equal to the natural logarithm of 𝑀 plus one.

We can now go back to the second part of our question, which asked us if the series in the question is convergent. We recall that we could determine the convergence or divergence of a series by looking at the limit of the partial sums. That is the sum from 𝑛 equals one to infinity of 𝑎 𝑛 is equal to the limit as 𝑛 approaches infinity of 𝑆 𝑛, where 𝑆 𝑛 is the 𝑛th partial sum of our series.

We already showed that the 𝑀th partial sum of our series is equal to the natural logarithm of 𝑀 plus one. So, to find the limit of our partial sums as 𝑀 approaches infinity is the same as calculating the limit of the natural logarithm of 𝑀 plus one as 𝑀 approaches infinity.

We recall that we know the limit of the natural logarithm function as 𝑥 approaches infinity is equal to infinity. So, we can just substitute 𝑥 is equal to 𝑀 plus one into this limit to give us the limit as 𝑀 plus one approaches infinity of the natural logarithm of 𝑀 plus one is equal to infinity. In particular, we know that if 𝑀 plus one is approaching infinity, then 𝑀 must also be approaching infinity. So, we can say our limit is as 𝑀 is approaching infinity.

Using this, we can conclude that the limit of our partial sums as 𝑀 approaches infinity is equal to infinity. Therefore, since the partial sums do not converge, we can conclude that the series must be divergent because the limit of the partial sums does not converge.