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I was able to prove this sum

$$\sum_{n=1}^\infty\frac{{1}}{n+3^n}$$

is convergent through the comparison test but I don't get how to find its sum.

David G. Stork
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Ford Davis
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  • This is such an interesting problem, but can you please provide more context? What class are you taking? – Toby Mak Dec 02 '18 at 00:15
  • I encountered a very similar situation a long time ago (I was asked to show it converged and was curious about its sum) and asked here. No answer was found, but you might find the comments interesting/useful. – Carl Schildkraut Dec 02 '18 at 00:16
  • Im in calc 2, my teacher didnt do a very good job of explaining this stuff and kinda rushed through the material since its the end of the semester. – Ford Davis Dec 02 '18 at 00:19
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    @FordDavis Ask your teacher about the question again. It's quite unlikely they expect for you to find the sum. Or maybe the question was $\frac{1}{n3^n}$, which has a much nicer answer. – Toby Mak Dec 02 '18 at 00:21
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    What about this one: Is there any non-zero function $f(n) $ such that $\sum_{n=1}^\infty \frac{1} {f(n) +3^n}$ has an okay closed form? – Zacky Dec 02 '18 at 00:22
  • Here is a picture of the assignment https://ibb.co/dsNbcnx – Ford Davis Dec 02 '18 at 00:27
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    @Zacky There are some trivial choices like $f(n) = 2^n - 3^n$. – D. Ungaretti Dec 02 '18 at 00:33
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    There's something odd about this assignment. Part b) can be summed in closed form, but it takes techniques way beyond calc 2. Are you perhaps supposed to approximate the sums. I definitely second @TobyMak's advice: ask your teacher about it. – saulspatz Dec 02 '18 at 00:33
  • @Daniel Or $f(n)=3^n$ – saulspatz Dec 02 '18 at 00:34
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    @FordDavis For the ones that converge, b) and g) look a little odd. However, d) is definitely solvable. – Toby Mak Dec 02 '18 at 00:39
  • Haha I overtought that, maybe I should've added non-trivial function $f(n) $. @Davis g) also looks unsolvable in a closed form. – Zacky Dec 02 '18 at 00:40
  • @saulspatz Part B) I actually asked her about yesterday because I couldnt figure it out and she told me to integrate it. I got [arctan(infinity)-arctan(1)]= pi/2 for that part. – Ford Davis Dec 02 '18 at 00:46
  • To get an idea about b) look here: https://math.stackexchange.com/q/1064217/515527 – Zacky Dec 02 '18 at 00:49
  • @Zacky the answer given there is way above my head, I barely even know what coth is. – Ford Davis Dec 02 '18 at 01:02
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    As noted by Carl, this question already exists ... https://math.stackexchange.com/q/1672121/442 but even that has no answer in nearly 3 years. No answer = cannot mark this as a duplicate. – GEdgar Dec 02 '18 at 01:14
  • Well, integrating it will given an approximation to the sum, but sums and integrals are different things. – saulspatz Dec 02 '18 at 03:03

1 Answers1

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What we can obtain, also by hand, is a reasonable estimation for example by

$$\sum_{n=1}^\infty\frac{{1}}{n+3^n}\approx 0.392<\sum_{n=1}^3\frac{{1}}{n+3^n}+\sum_{n=4}^\infty\frac{{1}}{3^n}=$$$$=\frac14+\frac1{11}+\frac1{30}+\frac32-1-\frac13-\frac1{9}-\frac1{27}\approx 0.393$$

user
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  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help. – Ford Davis Dec 02 '18 at 02:40
  • @FordDavis Yes indeed in that case seems convenient use the fact that $3^n$ is larger than $n$ also for $n$ small to obtain a god estimation for the series. – user Dec 02 '18 at 09:13