Finding the limit of a summation












0












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I have to find the limit as $n$ goes to infinity of the following expression-
$$sum_{r =1}^{n}frac{6n}{9n^2-r^2}$$



I tried to write the expression as $$frac{(3n+r)-(r-3n)}{(3n-r)(3n+r)}$$ and then separating the numerator in the hope of getting a telescoping series but that didn’t happen, can someone help me out.






calculus sequences-and-series limits







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  • $begingroup$
    Try Riemann sums.
    $endgroup$
    – xbh
    Jan 31 at 1:47
















0












$begingroup$


I have to find the limit as $n$ goes to infinity of the following expression-
$$sum_{r =1}^{n}frac{6n}{9n^2-r^2}$$



I tried to write the expression as $$frac{(3n+r)-(r-3n)}{(3n-r)(3n+r)}$$ and then separating the numerator in the hope of getting a telescoping series but that didn’t happen, can someone help me out.






calculus sequences-and-series limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try Riemann sums.
    $endgroup$
    – xbh
    Jan 31 at 1:47














0












0








0





$begingroup$


I have to find the limit as $n$ goes to infinity of the following expression-
$$sum_{r =1}^{n}frac{6n}{9n^2-r^2}$$



I tried to write the expression as $$frac{(3n+r)-(r-3n)}{(3n-r)(3n+r)}$$ and then separating the numerator in the hope of getting a telescoping series but that didn’t happen, can someone help me out.






calculus sequences-and-series limits







share|cite|improve this question









$endgroup$




I have to find the limit as $n$ goes to infinity of the following expression-
$$sum_{r =1}^{n}frac{6n}{9n^2-r^2}$$



I tried to write the expression as $$frac{(3n+r)-(r-3n)}{(3n-r)(3n+r)}$$ and then separating the numerator in the hope of getting a telescoping series but that didn’t happen, can someone help me out.






calculus sequences-and-series limits




calculus sequences-and-series limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 31 at 1:44









user601297user601297

41719




41719












  • $begingroup$
    Try Riemann sums.
    $endgroup$
    – xbh
    Jan 31 at 1:47


















  • $begingroup$
    Try Riemann sums.
    $endgroup$
    – xbh
    Jan 31 at 1:47
















$begingroup$
Try Riemann sums.
$endgroup$
– xbh
Jan 31 at 1:47




$begingroup$
Try Riemann sums.
$endgroup$
– xbh
Jan 31 at 1:47










2 Answers
2






active

oldest

votes


















3












$begingroup$

From Riemann sum, we have$$lim_{n to infty}sum_{r =1}^{n}frac{6n}{9n^2-r^2}=lim_{n to infty}frac1nsum_{r =1}^{n}frac{6}{9-left(frac{r}{n}right)^2}=int_0^1 frac{6}{9-x^2}, dx$$






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$endgroup$













  • $begingroup$
    Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
    $endgroup$
    – user601297
    Jan 31 at 1:51










  • $begingroup$
    hmmm can't think of other strategy for now.
    $endgroup$
    – Siong Thye Goh
    Jan 31 at 1:57



















1












$begingroup$

$$frac{6n}{9n^2-r^2} = frac{1}{3n-r} + frac{1}{3n+r}$$ so that $$S(n) = sum_{r=1}^n frac{6n}{9n^2-r^2} = -frac{1}{3n} + sum_{k=0}^{2n} frac{1}{2n+k} = -frac{1}{3n} + H_{4n} - H_{2n-1},$$ where $H_n$ is the $n^{rm th}$ harmonic number. Then as $n to infty$, $$H_n sim log n + gamma + mathcal O(1/n),$$ so $$lim_{n to infty} S(n) = lim_{n to infty} log 4n - log 2n + mathcal O(1/n) = log 2.$$






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  • $begingroup$
    Thanks for answering.
    $endgroup$
    – user601297
    Jan 31 at 2:05












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

From Riemann sum, we have$$lim_{n to infty}sum_{r =1}^{n}frac{6n}{9n^2-r^2}=lim_{n to infty}frac1nsum_{r =1}^{n}frac{6}{9-left(frac{r}{n}right)^2}=int_0^1 frac{6}{9-x^2}, dx$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
    $endgroup$
    – user601297
    Jan 31 at 1:51










  • $begingroup$
    hmmm can't think of other strategy for now.
    $endgroup$
    – Siong Thye Goh
    Jan 31 at 1:57
















3












$begingroup$

From Riemann sum, we have$$lim_{n to infty}sum_{r =1}^{n}frac{6n}{9n^2-r^2}=lim_{n to infty}frac1nsum_{r =1}^{n}frac{6}{9-left(frac{r}{n}right)^2}=int_0^1 frac{6}{9-x^2}, dx$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
    $endgroup$
    – user601297
    Jan 31 at 1:51










  • $begingroup$
    hmmm can't think of other strategy for now.
    $endgroup$
    – Siong Thye Goh
    Jan 31 at 1:57














3












3








3





$begingroup$

From Riemann sum, we have$$lim_{n to infty}sum_{r =1}^{n}frac{6n}{9n^2-r^2}=lim_{n to infty}frac1nsum_{r =1}^{n}frac{6}{9-left(frac{r}{n}right)^2}=int_0^1 frac{6}{9-x^2}, dx$$






share|cite|improve this answer









$endgroup$



From Riemann sum, we have$$lim_{n to infty}sum_{r =1}^{n}frac{6n}{9n^2-r^2}=lim_{n to infty}frac1nsum_{r =1}^{n}frac{6}{9-left(frac{r}{n}right)^2}=int_0^1 frac{6}{9-x^2}, dx$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 31 at 1:48









Siong Thye GohSiong Thye Goh

104k1468120




104k1468120












  • $begingroup$
    Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
    $endgroup$
    – user601297
    Jan 31 at 1:51










  • $begingroup$
    hmmm can't think of other strategy for now.
    $endgroup$
    – Siong Thye Goh
    Jan 31 at 1:57


















  • $begingroup$
    Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
    $endgroup$
    – user601297
    Jan 31 at 1:51










  • $begingroup$
    hmmm can't think of other strategy for now.
    $endgroup$
    – Siong Thye Goh
    Jan 31 at 1:57
















$begingroup$
Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
$endgroup$
– user601297
Jan 31 at 1:51




$begingroup$
Yeah i got the answer with this method, thanks a lot, out of curiosity, do you think there’s a way to solve this without Riemann sum?
$endgroup$
– user601297
Jan 31 at 1:51












$begingroup$
hmmm can't think of other strategy for now.
$endgroup$
– Siong Thye Goh
Jan 31 at 1:57




$begingroup$
hmmm can't think of other strategy for now.
$endgroup$
– Siong Thye Goh
Jan 31 at 1:57











1












$begingroup$

$$frac{6n}{9n^2-r^2} = frac{1}{3n-r} + frac{1}{3n+r}$$ so that $$S(n) = sum_{r=1}^n frac{6n}{9n^2-r^2} = -frac{1}{3n} + sum_{k=0}^{2n} frac{1}{2n+k} = -frac{1}{3n} + H_{4n} - H_{2n-1},$$ where $H_n$ is the $n^{rm th}$ harmonic number. Then as $n to infty$, $$H_n sim log n + gamma + mathcal O(1/n),$$ so $$lim_{n to infty} S(n) = lim_{n to infty} log 4n - log 2n + mathcal O(1/n) = log 2.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for answering.
    $endgroup$
    – user601297
    Jan 31 at 2:05
















1












$begingroup$

$$frac{6n}{9n^2-r^2} = frac{1}{3n-r} + frac{1}{3n+r}$$ so that $$S(n) = sum_{r=1}^n frac{6n}{9n^2-r^2} = -frac{1}{3n} + sum_{k=0}^{2n} frac{1}{2n+k} = -frac{1}{3n} + H_{4n} - H_{2n-1},$$ where $H_n$ is the $n^{rm th}$ harmonic number. Then as $n to infty$, $$H_n sim log n + gamma + mathcal O(1/n),$$ so $$lim_{n to infty} S(n) = lim_{n to infty} log 4n - log 2n + mathcal O(1/n) = log 2.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for answering.
    $endgroup$
    – user601297
    Jan 31 at 2:05














1












1








1





$begingroup$

$$frac{6n}{9n^2-r^2} = frac{1}{3n-r} + frac{1}{3n+r}$$ so that $$S(n) = sum_{r=1}^n frac{6n}{9n^2-r^2} = -frac{1}{3n} + sum_{k=0}^{2n} frac{1}{2n+k} = -frac{1}{3n} + H_{4n} - H_{2n-1},$$ where $H_n$ is the $n^{rm th}$ harmonic number. Then as $n to infty$, $$H_n sim log n + gamma + mathcal O(1/n),$$ so $$lim_{n to infty} S(n) = lim_{n to infty} log 4n - log 2n + mathcal O(1/n) = log 2.$$






share|cite|improve this answer









$endgroup$



$$frac{6n}{9n^2-r^2} = frac{1}{3n-r} + frac{1}{3n+r}$$ so that $$S(n) = sum_{r=1}^n frac{6n}{9n^2-r^2} = -frac{1}{3n} + sum_{k=0}^{2n} frac{1}{2n+k} = -frac{1}{3n} + H_{4n} - H_{2n-1},$$ where $H_n$ is the $n^{rm th}$ harmonic number. Then as $n to infty$, $$H_n sim log n + gamma + mathcal O(1/n),$$ so $$lim_{n to infty} S(n) = lim_{n to infty} log 4n - log 2n + mathcal O(1/n) = log 2.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 31 at 2:04









heropupheropup

65.3k865104




65.3k865104












  • $begingroup$
    Thanks for answering.
    $endgroup$
    – user601297
    Jan 31 at 2:05


















  • $begingroup$
    Thanks for answering.
    $endgroup$
    – user601297
    Jan 31 at 2:05
















$begingroup$
Thanks for answering.
$endgroup$
– user601297
Jan 31 at 2:05




$begingroup$
Thanks for answering.
$endgroup$
– user601297
Jan 31 at 2:05


















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