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Find all positive values of $ b $ for which the series $ \sum_{n = 1}^{\infty} b^{\ln n} $ converges.

since $b$ has to be positive also, so $\sum_{n=1}^{\infty} b^{\ln n}$ will converge as long as $0<$

$b<\frac{1}{e}$

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Missouri State University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

in this problem, we'd like to find all the values of B there allow the Siri's to converge. So what we'LL do is just start off by playing with this and term here. So recall that weaken right b is e to the Ellen B. So rewrite bee in this form and then properties of exponents says when you raise a power to another power you got it won't supply those powers. So here, let me go ahead. And actually, and because of that fact, let me just go ahead and change the role of and then B And the reason again is because of this fact Here, thes two quantities are equal because they're both equal to this term here. So here I am, just switching the exponents using this fact. And then now, using this again, I can rewrite the term in the print. Decease is just end. And then I get n Ellen be and we let's write This is one over and negative. Ellen, be so we've shown here Is that the original Siri's Khun B. Ren as a P series and this will make it easier for us to answer the question. So this is Ahh P Siri's with P negative ln b and remember, p series converges if and only if if peace is bigger than one. So we need negative Ellen be to be larger than one and then we'LL just solve this for being Some will supply both sides by negative one and take e on both sides. The reason this is true is because if X is less than why than e X is less than either the y so that justifies this fact here. And then we can rewrite the left hand side again by using This is B and so we have B is less than one over he So here we wanted also we wanted positive values. If you go back to the original wording in the problem, so here we should also throw in the condition that these larger than zero hey, and that's your final answer