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Evaluate : lim(n-> oo) (1^4+2^4+3^4+...+...

Evaluate : `lim_(n-> oo) (1^4+2^4+3^4+...+n^4)/n^5 - lim_(n->oo) (1^3+2^3+...+n^3)/n^5`

A

`1/30`

B

zero

C

`1/4`

D

`1/5`

Text Solution

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The correct Answer is:
To evaluate the expression \[ \lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + \ldots + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + \ldots + n^3}{n^5}, \] we will use the standard formulas for the sums of powers of integers. ### Step 1: Evaluate the first limit The sum of the first \( n \) natural numbers raised to the fourth power is given by the formula: \[ \sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}. \] Substituting this into the limit, we have: \[ \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^4}{n^5} = \lim_{n \to \infty} \frac{\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}}{n^5}. \] ### Step 2: Simplify the first limit This simplifies to: \[ \lim_{n \to \infty} \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30n^5}. \] Dividing the numerator and denominator by \( n^5 \): \[ = \lim_{n \to \infty} \frac{(1)(1 + \frac{1}{n})(2 + \frac{1}{n})\left(3 + \frac{3}{n} - \frac{1}{n^2}\right)}{30}. \] ### Step 3: Evaluate the limit As \( n \to \infty \), the terms \( \frac{1}{n} \) and \( \frac{1}{n^2} \) approach 0. Thus, we have: \[ = \frac{1 \cdot 1 \cdot 2 \cdot 3}{30} = \frac{6}{30} = \frac{1}{5}. \] ### Step 4: Evaluate the second limit The sum of the first \( n \) natural numbers raised to the third power is given by the formula: \[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \] Substituting this into the limit, we have: \[ \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^3}{n^5} = \lim_{n \to \infty} \frac{\left(\frac{n(n+1)}{2}\right)^2}{n^5}. \] ### Step 5: Simplify the second limit This simplifies to: \[ = \lim_{n \to \infty} \frac{n^2(n+1)^2}{4n^5} = \lim_{n \to \infty} \frac{(1)(1 + \frac{1}{n})^2}{4}. \] ### Step 6: Evaluate the limit As \( n \to \infty \), the term \( \frac{1}{n} \) approaches 0. Thus, we have: \[ = \frac{1 \cdot 1^2}{4} = \frac{1}{4}. \] ### Step 7: Combine the results Now we can combine the results from the two limits: \[ \frac{1}{5} - \frac{1}{4}. \] To combine these fractions, we find a common denominator, which is 20: \[ = \frac{4}{20} - \frac{5}{20} = -\frac{1}{20}. \] ### Final Answer Thus, the final answer is: \[ \boxed{-\frac{1}{20}}. \]
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VIKAS GUPTA (BLACK BOOK) ENGLISH-INDEFINITE AND DEFINITE INTEGRATION-EXERCISE (SUBJECTIVE TYPE PROBLEMS)
  1. Evaluate : lim(n-> oo) (1^4+2^4+3^4+...+n^4)/n^5 - lim(n->oo) (1^3+2^3...

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  2. int (x+ ( cos^(-1)3x )^(2))/(sqrt(1-9x ^(2)))dx = (1)/(k (1)) ( sqrt(1...

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  3. If int (0)^(oo) (x ^(3))/((a ^(2)+ x ^(2)))dx = (1)/(ka ^(6)), then fi...

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  4. Let f (x) = x cos x, x in [(3pi)/(2), 2pi] and g (x) be its inverse. ...

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  5. If int (x ^(6) +x ^(4)+x^(2)) sqrt(2x ^(4) +3x ^(2)+6) dx = ((ax ^(6) ...

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  6. If the value of the definite integral int ^(-1) ^(1) cos ^(-1) ((1)/(s...

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  7. The value of int (tan x )/(tan ^(2) x + tan x+1)dx =x -(2)/(sqrtA) tan...

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  8. Let int (0)^(1) (4x ^(3) (1+(x ^(4)) ^(2010)))/((1+x^(4))^(2012))dx = ...

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  9. Let int ( 1) ^(sqrt5)(x ^(2x ^(2)+1) +ln "("x ^(2x ^(2x ^(2)+1))")")dx...

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  10. If int (dx )/(cos ^(3) x-sin ^(3))=A tan ^(-1) (f (x)) +bln |(sqrt2+f ...

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  11. Find the value of |a| for which the area of triangle included between ...

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  12. Let I = int (0) ^(pi) x ^(6) (pi-x) ^(8)dx, then (pi ^(15))/((""^(15) ...

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  13. If I = int (0) ^(100) (sqrtx)dx, then the value of (9I)/(155) is:

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  14. Let I(n) = int (0)^(pi) (sin (n + (1)/(2))x )/(sin ((x)/(2)))dx where ...

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  15. If M be the maximum value of 72 int (0) ^(y) sqrt(x ^(4) +(y-y^(2))^(2...

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  16. Find the number points where f (theta) = int (-1)^(1) (sin theta dx )/...

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  17. underset(nrarroo)lim[(1)/(sqrtn)+(1)/(sqrt(2n))+(1)/(sqrt(3n))+...+(1)...

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  18. The maximum value of int (-pi//2) ^(2pi//2) sin x. f (x) dx, subject t...

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  19. Given a function g, continous everywhere such that g (1)=5 and int (0)...

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  20. If f (n)= 1/pi int (0) ^(pi//2) (sin ^(2) (n theta) d theta)/(sin ^(2)...

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  21. Let f (2-x) =f (2+xand f (4-x )= f (4+x). Function f (x) satisfies int...

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