If l-limn-nr-2r-1sin-r-1-rsin-r then find l- where denotes the fractional part function-

l=lim _ n→∞ #160; ∑ _ r=2 ^ n ( ( r+1 ) sinπ #160; ( r+1 ) #160; rsinπ #160; r #160; #160; #160; ) #160; #160; ∑ _ r=2 ^ n ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

If l=limn→∞...

Question

If $l = lim n \to \infty n \sum r = 2 ((r + 1) sin \frac{π}{r + 1} - r sin \frac{π}{r})$ then find ${l}$ . (where ${}$ denotes the fractional part function.

Open in App

Solution

$l = lim n \to \infty n \sum r = 2 ((r + 1) sin \frac{π}{(r + 1)} - r sin \frac{π}{r})$

$\sum_{r = 2}^{n} ((r + 1) sin \frac{π}{(r + 1)} - r sin \frac{π}{r})$

$= 3 sin \frac{π}{3} - 2 sin \frac{π}{2} + 4 sin \frac{π}{4} - 3 sin \frac{π}{3} + . . . + (n + 1) sin \frac{π}{n + 1} - n sin \frac{π}{n}$

$= (n + 1) sin \frac{π}{(n + 1)} - 2$
$l = lim n \to \infty ⎛ ⎜ ⎜ ⎜ ⎝ \frac{sin \frac{π}{n + 1}}{\frac{1}{n + 1} \times \frac{π}{π}} - 2 ⎞ ⎟ ⎟ ⎟ ⎠$

as $n \to \infty \frac{1}{n + 1} \to 0; \frac{π}{n + 1} \to 0$

$= π \times 1 - 2 = \frac{22}{7} - 2 = \frac{8}{7}$
${l} = {\frac{8}{7}} = \frac{1}{7}$

Suggest Corrections

Similar questions

Q. Evaluate

lim n \to \infty n \sum r = 1 \frac{1}{2^{r}}

where

[.]

denotes the greatest integer function.

lim x \to 0^{-} {{sin}^{- 1} [tan x]} = l

then

l

is equal to
where

[]

and

{}

denotes greatest the integer and fraction part function

If $f (x) = l i m n \to \infty \sum_{r - = 1}^{n} \frac{r x}{1.3.5 \dots (2 r + 1)}$ , then $\int_{0}^{3} [f (x)] d (x - [x])$ is equal to (where [.] denotes greatest integer function)

Q. If

l, m

and

n

are number of points of discontinuity, non-differentiability and local extrema of function

f (x) = max [\sqrt{1 - x^{2}}, {x}]

x \in [- 1, 1]

respectively then

l + m + n

is equal to
[where

{\cdot}

denotes fractional part function]

Q. If

S_{n} = n \sum r = 1 r - 1 \sum t = 0 (\frac{1}{6^{n}}^{n} C_{r}^{r} C_{t} 4^{t}),

then the value of

l

where

l = \infty \sum n = 1 (1 - S_{n})

Join BYJU'S Learning Program

If l=limn→∞n∑r=2((r+1)sinπr+1−rsinπr) then find {l}. (where {} denotes the fractional part function.

l=limn→∞n∑r=2((r+1)sinπ(r+1)−rsinπr)∑nr=2((r+1)sinπ(r+1)−rsinπr)=3sinπ3−2sinπ2+4sinπ4−3sinπ3+...+(n+1)sinπn+1−nsinπn=(n+1)sinπ(n+1)−2l=limn→∞⎛⎜ ⎜ ⎜⎝sinπn+11n+1×ππ−2⎞⎟ ⎟ ⎟⎠as n→∞1n+1→0;πn+1→0=π×1−2=227−2=87{l}={87}=17

If $l = lim n \to \infty n \sum r = 2 ((r + 1) sin \frac{π}{r + 1} - r sin \frac{π}{r})$ then find ${l}$ . (where ${}$ denotes the fractional part function.