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# Is the 50th partial sum $s_{50}$ of the alternating series $\displaystyle \sum_{n = 1}^{\infty} (-1)^{n - 1} /n$ an overestimate or an underestimate of the total sum? Explain.

## underestimate since $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=s_{50}+\left(\frac{1}{51}-\frac{1}{52}\right)+\left(\frac{1}{53}-\frac{1}{54}\right)+\cdots \cdot,$ and the terms in parentheses are all positive.

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is the fiftieth. Partial some and overestimate or underestimate for the total sum. So for all trading theories, we always have one of the following situations. So the even sub scripts are always less than the sum, which is always less than the odd sub scripts or the opposite is true. This is shrewd and saw the picture of each case. Here's your some and then you have, like, a swan three as five and so on. That's not going in the right direction. So let me go back and change that. Here's s one as three and so on. And then here you have us, too. As for oranges, have things what switched around a little bit as one history. So they're getting closer to us. But all the evens are on the right and all the odds are on the left. So here to find out what case we're in. So case Air Case B. So it's one or the other, not both. So let's find out which case we're in. So let's find that one as one is just one, and this will be an overestimate since, for example, we can just the way to answer this the way to find out whether it's an overestimate, as one I'm saying is just by comparing it test too, so as to is s one plus and then we have negative one to the two minus one over two. So that's one minus a half much is a half, so we can see that as one is bigger than as too. So we must be in case, eh? It's one or the other. So therefore, as one is overestimate. So this tells us that all the odds are in our overestimate. So the even numbers are the lower estimates. So since ESA fifty will fall on this side, fifty is even so as the Sun fifty is less than the entire sum. So that's not a five that's s there. So we don't know that with with a dollar sign to distinguish my my s is from fives. So that's a sum from one to infinity minus one to the end, minus one over end. So we see that this is an underestimate, since it's less than and that's our final answer

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