How to calculate the sum of $sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)$












1














I know this series is converges by limit test, but i can't find a way to calculate its sum.



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)$$
Thanks for helping,Here is the solution.



Solutions :
$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$
First thing to do is taking out the constant terms, which is $$left(frac45right)^{-1},left(frac15right)^{-1}$$
We can see there are two geometric series:
$$sumlimits_{n=1}^{infty}left(frac45right)^{n}=frac{frac45}{1-frac45}=4$$
Another one is: $$sumlimits_{n=1}^{infty}left(frac15right)^{n}=frac{frac45}{1-frac45}=frac14$$



So,we have
$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=frac{5}{4}cdot4+4cdot5cdotfrac14$$



$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=10$$






convergence







share|cite|improve this question
























  • You should add some detail more about the context for your answer and what you have tried with that.
    – gimusi
    Dec 12 '18 at 15:52










  • Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
    – gimusi
    Dec 13 '18 at 17:56
















1














I know this series is converges by limit test, but i can't find a way to calculate its sum.



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)$$
Thanks for helping,Here is the solution.



Solutions :
$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$
First thing to do is taking out the constant terms, which is $$left(frac45right)^{-1},left(frac15right)^{-1}$$
We can see there are two geometric series:
$$sumlimits_{n=1}^{infty}left(frac45right)^{n}=frac{frac45}{1-frac45}=4$$
Another one is: $$sumlimits_{n=1}^{infty}left(frac15right)^{n}=frac{frac45}{1-frac45}=frac14$$



So,we have
$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=frac{5}{4}cdot4+4cdot5cdotfrac14$$



$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=10$$






convergence







share|cite|improve this question
























  • You should add some detail more about the context for your answer and what you have tried with that.
    – gimusi
    Dec 12 '18 at 15:52










  • Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
    – gimusi
    Dec 13 '18 at 17:56














1












1








1







I know this series is converges by limit test, but i can't find a way to calculate its sum.



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)$$
Thanks for helping,Here is the solution.



Solutions :
$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$
First thing to do is taking out the constant terms, which is $$left(frac45right)^{-1},left(frac15right)^{-1}$$
We can see there are two geometric series:
$$sumlimits_{n=1}^{infty}left(frac45right)^{n}=frac{frac45}{1-frac45}=4$$
Another one is: $$sumlimits_{n=1}^{infty}left(frac15right)^{n}=frac{frac45}{1-frac45}=frac14$$



So,we have
$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=frac{5}{4}cdot4+4cdot5cdotfrac14$$



$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=10$$






convergence







share|cite|improve this question















I know this series is converges by limit test, but i can't find a way to calculate its sum.



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)$$
Thanks for helping,Here is the solution.



Solutions :
$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$
First thing to do is taking out the constant terms, which is $$left(frac45right)^{-1},left(frac15right)^{-1}$$
We can see there are two geometric series:
$$sumlimits_{n=1}^{infty}left(frac45right)^{n}=frac{frac45}{1-frac45}=4$$
Another one is: $$sumlimits_{n=1}^{infty}left(frac15right)^{n}=frac{frac45}{1-frac45}=frac14$$



So,we have
$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=frac{5}{4}cdot4+4cdot5cdotfrac14$$



$$sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}=10$$






convergence




convergence






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share|cite|improve this question













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edited Dec 13 '18 at 8:25









rtybase

10.5k21533




10.5k21533










asked Dec 12 '18 at 15:44









PatricEzPatricEz

92




92












  • You should add some detail more about the context for your answer and what you have tried with that.
    – gimusi
    Dec 12 '18 at 15:52










  • Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
    – gimusi
    Dec 13 '18 at 17:56


















  • You should add some detail more about the context for your answer and what you have tried with that.
    – gimusi
    Dec 12 '18 at 15:52










  • Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
    – gimusi
    Dec 13 '18 at 17:56
















You should add some detail more about the context for your answer and what you have tried with that.
– gimusi
Dec 12 '18 at 15:52




You should add some detail more about the context for your answer and what you have tried with that.
– gimusi
Dec 12 '18 at 15:52












Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
– gimusi
Dec 13 '18 at 17:56




Very nice work! Now as you can see you question has been reopened after your improvements. Please for the next time follow the indications given here on How to ask a good question. Bye
– gimusi
Dec 13 '18 at 17:56










1 Answer
1






active

oldest

votes


















2














We can use that



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$



Can you conclude using geometric series?






share|cite|improve this answer





















  • I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
    – PatricEz
    Dec 12 '18 at 15:57










  • Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
    – gimusi
    Dec 12 '18 at 15:59










  • I will add it in my question.Thanks for your help :P
    – PatricEz
    Dec 12 '18 at 16:01










  • @PatricEz You are welcome! Bye
    – gimusi
    Dec 12 '18 at 16:01











Your Answer





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1 Answer
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1 Answer
1






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oldest

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active

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2














We can use that



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$



Can you conclude using geometric series?






share|cite|improve this answer





















  • I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
    – PatricEz
    Dec 12 '18 at 15:57










  • Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
    – gimusi
    Dec 12 '18 at 15:59










  • I will add it in my question.Thanks for your help :P
    – PatricEz
    Dec 12 '18 at 16:01










  • @PatricEz You are welcome! Bye
    – gimusi
    Dec 12 '18 at 16:01
















2














We can use that



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$



Can you conclude using geometric series?






share|cite|improve this answer





















  • I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
    – PatricEz
    Dec 12 '18 at 15:57










  • Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
    – gimusi
    Dec 12 '18 at 15:59










  • I will add it in my question.Thanks for your help :P
    – PatricEz
    Dec 12 '18 at 16:01










  • @PatricEz You are welcome! Bye
    – gimusi
    Dec 12 '18 at 16:01














2












2








2






We can use that



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$



Can you conclude using geometric series?






share|cite|improve this answer












We can use that



$$sumlimits_{n=1}^{infty}left(frac{4^{n-1}}{5^{n-1}}+frac{4}{5^{n-1}}right)=sumlimits_{n=1}^{infty}left(frac45right)^{n-1}+4sumlimits_{n=1}^{infty}left(frac15right)^{n-1}$$



Can you conclude using geometric series?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 15:46









gimusigimusi

1




1












  • I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
    – PatricEz
    Dec 12 '18 at 15:57










  • Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
    – gimusi
    Dec 12 '18 at 15:59










  • I will add it in my question.Thanks for your help :P
    – PatricEz
    Dec 12 '18 at 16:01










  • @PatricEz You are welcome! Bye
    – gimusi
    Dec 12 '18 at 16:01


















  • I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
    – PatricEz
    Dec 12 '18 at 15:57










  • Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
    – gimusi
    Dec 12 '18 at 15:59










  • I will add it in my question.Thanks for your help :P
    – PatricEz
    Dec 12 '18 at 16:01










  • @PatricEz You are welcome! Bye
    – gimusi
    Dec 12 '18 at 16:01
















I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
– PatricEz
Dec 12 '18 at 15:57




I see it now...two geometric series,so it becomes (5/4)*((4/5)/(1-(4/5)))+4*((1/5)/(1-(1/5)))=10.Am i right? Thanks
– PatricEz
Dec 12 '18 at 15:57












Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
– gimusi
Dec 12 '18 at 15:59




Yes exactly! Well done. It would be nice if you can add your solution editing your question. Bye
– gimusi
Dec 12 '18 at 15:59












I will add it in my question.Thanks for your help :P
– PatricEz
Dec 12 '18 at 16:01




I will add it in my question.Thanks for your help :P
– PatricEz
Dec 12 '18 at 16:01












@PatricEz You are welcome! Bye
– gimusi
Dec 12 '18 at 16:01




@PatricEz You are welcome! Bye
– gimusi
Dec 12 '18 at 16:01


















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