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Question Number 45155 by rahul 19 last updated on 09/Oct/18

Commented by rahul 19 last updated on 09/Oct/18

${I t}^{'} s a v e r y b e a u t i f u l Q u e s . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Oct/18

	$S_{n} = \underset{n}{\sum^{n}} \frac{r^{4}}{n^{5}} + \frac{r^{3} n}{n^{5}} + \frac{r^{2} n^{2}}{n^{5}} + \frac{2 n^{4}}{n^{5}}$ $= \frac{1}{n^{5}} \underset{r = 1}{\sum^{n}} r^{4} + \frac{1}{n^{4}} \underset{r = 1}{\sum^{n}} r^{3} + \frac{1}{n^{3}} \underset{r = 1}{\sum^{n}} r^{2} + \frac{2}{n} \underset{r = 1}{\sum^{n}} 1$ $= \frac{1}{n^{5}} (\frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}) + \frac{1}{n^{4}} (\frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}) + \frac{1}{n^{3}} (\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}) + \frac{2}{n} (n)$ $= \frac{1}{5} + \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{30 n^{4}} + \frac{1}{4} + \frac{1}{2 n} + \frac{1}{4 n^{2}} + \frac{1}{3} + \frac{1}{2 n} + \frac{1}{6 n^{2}} + 2$ $= (\frac{1}{5} + \frac{1}{4} + \frac{1}{3} + 2) + f (n)$ $= \frac{167}{60} + f (n)$ $S_{n} > \frac{167}{60}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Oct/18
	$T_n =(1/n^5 )Σ_(r=0) ^(n−1) r^4 +(1/n^4 )Σ_(r=0) ^(n−1) r^3 +(1/n^3 )Σ_(r=0) ^(n−1) r^2 +(2/n)Σ_(r=0) ^(n−1) 1 ={(1/n^5 )Σ_(r=1) ^n r^4 +(1/n^4 )Σ_(r=1) ^n r^3 +(1/n^3 )Σ_(r=1) ^n r^2 +(2/n)Σ_(r=1) ^n }−{(n^4 /n^5 )+(n^3 /n^4 )+(n^2 /n^3 )} =((167)/(60))+f(n)−{(1/n)+(1/n)+(1/n)} =((167)/(60))+((3/(2n))+(3/(4n^2 ))−(1/(30n^4 )))−(3/n) =((167)/(60))+((3/(4n^2 ))−(1/(30n^4 ))−(3/(2n))) =((167)/(60))+(((45n^2 −2−90n^3 )/(60n^4 ))) =((167)/(60))+((45n^2 (1−n)−2)/(60n^4 )) =((167)/(60))−{((45n^2 (n−1)+2)/(60n^4 ))} so T_n <((167)/(60)) value of f(n) found in S_n f(n)=(1/(2n))+(1/(3n^2 ))−(1/(30n^4 ))+(1/(2n))+(1/(4n^2 ))+(1/(2n))+(1/(6n^2 )) =((3/(2n))+(9/(12n^2 ))−(1/(30n^4 )))$
	$T_{n} = \frac{1}{n^{5}} \underset{r = 0}{\sum^{n - 1}} r^{4} + \frac{1}{n^{4}} \underset{r = 0}{\sum^{n - 1}} r^{3} + \frac{1}{n^{3}} \underset{r = 0}{\sum^{n - 1}} r^{2} + \frac{2}{n} \underset{r = 0}{\sum^{n - 1}} 1$ $= {\frac{1}{n^{5}} \underset{r = 1}{\sum^{n}} r^{4} + \frac{1}{n^{4}} \underset{r = 1}{\sum^{n}} r^{3} + \frac{1}{n^{3}} \underset{r = 1}{\sum^{n}} r^{2} + \frac{2}{n} \underset{r = 1}{\sum^{n}}} - {\frac{n^{4}}{n^{5}} + \frac{n^{3}}{n^{4}} + \frac{n^{2}}{n^{3}}}$ $= \frac{167}{60} + f (n) - {\frac{1}{n} + \frac{1}{n} + \frac{1}{n}}$ $= \frac{167}{60} + (\frac{3}{2 n} + \frac{3}{4 n^{2}} - \frac{1}{30 n^{4}}) - \frac{3}{n}$ $= \frac{167}{60} + (\frac{3}{4 n^{2}} - \frac{1}{30 n^{4}} - \frac{3}{2 n})$ $= \frac{167}{60} + (\frac{45 n^{2} - 2 - 90 n^{3}}{60 n^{4}})$ $= \frac{167}{60} + \frac{45 n^{2} (1 - n) - 2}{60 n^{4}}$ $= \frac{167}{60} - {\frac{45 n^{2} (n - 1) + 2}{60 n^{4}}}$ $s o T_{n} < \frac{167}{60}$ $v a l u e o f f (n) f o u n d i n S_{n}$ $f (n) = \frac{1}{2 n} + \frac{1}{3 n^{2}} - \frac{1}{30 n^{4}} + \frac{1}{2 n} + \frac{1}{4 n^{2}} + \frac{1}{2 n} + \frac{1}{6 n^{2}}$ $= (\frac{3}{2 n} + \frac{9}{12 n^{2}} - \frac{1}{30 n^{4}})$
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