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# Determine whether the series is convergent or divergent. If it is convergent, find its sum.$$\displaystyle \sum_{k = 0}^{\infty} (\sqrt 2)^{-k}$$

## $\frac{\sqrt{2}}{\sqrt{2}-1}$

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let's determine whether or not this series converges. And then if it does converge, we'll go ahead and find the sum. So this looks almost like a geometric series, except we have a negative cater. So let's go ahead and just rewrite this K equals zero infinity. Tell me, write this as one over the square root of two, all to the K power. And now we can see that this is geometric. So recall geometric means that you have a series and could start at zero. We can start at one. Doesn't matter here. What? The starting point is what matters is that it has this form here where a and are are just numbers they don't depend on end. All right, so we see this We c r r equals one over square root of two and recall that since our this, our value is less than one. The geometric series will converge. All right, so this series is convergent. That's our first answer up here. But since we have conversion as an answer, we got some more work to do. Let's go and find that some. So you have a formula for geometric series, so let's go ahead and use that. So this equals the formula to use. In any case, for geometric, it's always the first term of the series, which depends on the symbol right here and the denominator or underneath the sigma. That first number is the first tells you how to get the first term and then one minus R. So in our problem, the first term is when you plug in K equals zero, you just get a one. And then So let me even write that I could show more work there squared of two to the negative zero, which is just 1/1 minus R. So I just have one up top and then route to minus one over root two, which is just route to over route to minus one. And if you'd like, you could go ahead and rationalize the denominator. But otherwise there's our final answer. So it's conversion, and this is the value of the sum

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