Calculating $sumlimits_{n=1}^inftyfrac{{1}}{n+3^n} $












8














I was able to prove this sum



$$sum_{n=1}^inftyfrac{{1}}{n+3^n}$$



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






calculus sequences-and-series convergence summation







share|cite|improve this question




















  • 2




    @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 2 at 0:21








  • 2




    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 2 at 0:22








  • 1




    @Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
    – Daniel
    Dec 2 at 0:33






  • 1




    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 2 at 0:33






  • 3




    As noted by Carl, this question already exists ... 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 2 at 1:14


















8














I was able to prove this sum



$$sum_{n=1}^inftyfrac{{1}}{n+3^n}$$



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






calculus sequences-and-series convergence summation







share|cite|improve this question




















  • 2




    @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 2 at 0:21








  • 2




    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 2 at 0:22








  • 1




    @Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
    – Daniel
    Dec 2 at 0:33






  • 1




    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 2 at 0:33






  • 3




    As noted by Carl, this question already exists ... 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 2 at 1:14
















8












8








8


4





I was able to prove this sum



$$sum_{n=1}^inftyfrac{{1}}{n+3^n}$$



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






calculus sequences-and-series convergence summation







share|cite|improve this question















I was able to prove this sum



$$sum_{n=1}^inftyfrac{{1}}{n+3^n}$$



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






calculus sequences-and-series convergence summation




calculus sequences-and-series convergence summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 at 0:51









David G. Stork

9,73921232




9,73921232










asked Dec 2 at 0:05









Ford Davis

493




493








  • 2




    @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 2 at 0:21








  • 2




    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 2 at 0:22








  • 1




    @Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
    – Daniel
    Dec 2 at 0:33






  • 1




    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 2 at 0:33






  • 3




    As noted by Carl, this question already exists ... 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 2 at 1:14
















  • 2




    @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 2 at 0:21








  • 2




    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 2 at 0:22








  • 1




    @Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
    – Daniel
    Dec 2 at 0:33






  • 1




    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 2 at 0:33






  • 3




    As noted by Carl, this question already exists ... 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 2 at 1:14










2




2




@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 2 at 0:21






@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 2 at 0:21






2




2




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 2 at 0:22






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 2 at 0:22






1




1




@Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
– Daniel
Dec 2 at 0:33




@Zacky There are some trivial choices like $f(n) = 2^n - 3^n$.
– Daniel
Dec 2 at 0:33




1




1




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 2 at 0:33




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 2 at 0:33




3




3




As noted by Carl, this question already exists ... 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 2 at 1:14






As noted by Carl, this question already exists ... 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 2 at 1:14












1 Answer
1






active

oldest

votes


















3














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



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






share|cite|improve this answer























  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
    – Ford Davis
    Dec 2 at 2:40










  • @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.
    – gimusi
    Dec 2 at 9:13











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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



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






share|cite|improve this answer























  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
    – Ford Davis
    Dec 2 at 2:40










  • @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.
    – gimusi
    Dec 2 at 9:13
















3














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



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






share|cite|improve this answer























  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
    – Ford Davis
    Dec 2 at 2:40










  • @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.
    – gimusi
    Dec 2 at 9:13














3












3








3






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



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






share|cite|improve this answer














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



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 at 9:14

























answered Dec 2 at 2:24









gimusi

1




1












  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
    – Ford Davis
    Dec 2 at 2:40










  • @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.
    – gimusi
    Dec 2 at 9:13


















  • I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
    – Ford Davis
    Dec 2 at 2:40










  • @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.
    – gimusi
    Dec 2 at 9:13
















I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
– Ford Davis
Dec 2 at 2:40




I dont remember us going over approximating the sum, but this might be what she wanted. Thanks for your help.
– Ford Davis
Dec 2 at 2:40












@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.
– gimusi
Dec 2 at 9:13




@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.
– gimusi
Dec 2 at 9:13


















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