Video: Determine the Partial Sum of a Series of Inverse Trigonometric Functions

Find the partial sum of the series βˆ‘_(𝑛 = 1)^(∞) tan⁻¹ (𝑛 + 1) βˆ’ tan⁻¹ (𝑛).

02:51

Video Transcript

Find the partial sum of the series the sum from 𝑛 equals one to ∞ of the inverse tan of 𝑛 plus one minus the inverse tan of 𝑛.

The question gives us an infinite series. It wants us to find an expression for the partial sum of this series. We recall we call 𝑠 sub 𝑛 the partial sum of a series. It’s the sum of the first 𝑛 terms of the series. So in our case, our 𝑛th partial sum 𝑠 sub 𝑛 will be the sum from 𝑖 equals one to 𝑛 of the inverse tan of 𝑖 plus one minus the inverse tan of 𝑖.

To help us find an expression for 𝑠 sub 𝑛, let’s expand our series. We’ll work it out term by term. Let’s start by finding the first term in our series. That’s when 𝑖 is equal to one. We get the inverse tan of one plus one minus the inverse tan of one. Next, we’ll simplify this expression slightly. One plus one is equal to two. So the first term simplifies to give us the inverse tan of two plus the inverse tan of one.

Let’s now add on the second term in our series. That’s when 𝑖 is equal to two. Substituting 𝑖 is equal to two, we get the inverse tan of two plus one minus the inverse tan of two. Now, we can see something interesting. We’re adding the inverse tan of two and then subtracting it. In fact, if we keep adding more terms, we’ll find another useful property.

Our third term will be the inverse tan to three plus one minus the inverse tan of three. And we can now see the inverse tan of two plus one minus the inverse tan of three is equal to zero. So we can see something useful. Every time we add a new term to our series, it cancels with part of the series that came before it. So what would happen at the end of our partial sum? Let’s add all the terms.

We’ll write out in full the last two terms of our partial sum. These are the inverse tan of 𝑛 minus one plus one minus the inverse tan of 𝑛 minus one plus the inverse tan of 𝑛 plus one minus the inverse tan of 𝑛. Remember, when we add a term to our series, the second part of that term cancel with part of the series which came before it. So negative the inverse tan of 𝑛 minus one canceled with part of the series that came before.

Next, we can cancel the inverse tan of 𝑛 minus one plus one with negative the inverse tan of 𝑛. Finally, if we were to include the fourth term in our expansion, we would have negative the inverse tan of four. So we can cancel the inverse tan of three plus one with this. So in actual fact, everything cancels except for two terms. We have negative the inverse tan of one plus the inverse tan of 𝑛 plus one. We call this property a telescoping series.

And when we order these two terms. Therefore, we’ve shown the partial sum 𝑠 sub 𝑛 of the series the sum from 𝑛 equals one to ∞ of the inverse tan of 𝑛 plus one minus the inverse tan of 𝑛 is equal to the inverse tan of 𝑛 plus one minus the inverse tan of one.

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