Question

The value of the limit
\(lim _{n→∞}(\frac{1^4+2^4+…+n^4}{n^5}+\frac{1}{√n}(\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+...+\frac{1}{\sqrt{4n}}))\)
is equal to___(Rounded off two decimal places)

Answer 1

Correct Answer: 2.19 - 2.21

The problem asks to evaluate the limit:

$$L = \lim_{n \to \infty} \left( \underbrace{\frac{1^4 + 2^4 + \dots + n^4}{n^5}}_{L_1} + \underbrace{\frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \dots + \frac{1}{\sqrt{4n}} \right)}_{L_2} \right)$$

We will solve this by splitting it into two parts, $L_1$ and $L_2$, evaluating them separately using the concept of Definite Integral as the Limit of a Sum.

The general formula used is:

$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=a}^{bn} f\left(\frac{r}{n}\right) = \int_{a/n}^{bn/n} f(x) \, dx$$

Part 1: Evaluate $L_1$

$$L_1 = \lim_{n \to \infty} \frac{1^4 + 2^4 + \dots + n^4}{n^5}$$

Rewrite the sum using sigma notation:

 

$$L_1 = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r^4}{n^5}$$

Separate $\frac{1}{n}$ to form the Riemann sum structure:

 

$$L_1 = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^4$$

Here, $f(x) = x^4$. The lower limit is $\frac{r}{n} \to 0$ (as $r=1, n \to \infty$) and the upper limit is $\frac{r}{n} \to 1$ (as $r=n$).

 

$$L_1 = \int_{0}^{1} x^4 \, dx$$

Now, integrate:

$$L_1 = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5} - 0 = 0.2$$

So, $L_1 = 0.2$

Part 2: Evaluate $L_2$

$$L_2 = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \dots + \frac{1}{\sqrt{4n}} \right)$$

Rewrite the series in sigma notation. Note that the last term is $\frac{1}{\sqrt{4n}}$, which can be written as $\frac{1}{\sqrt{n+3n}}$. This means the summation index $r$ goes from $1$ to $3n$.

$$L_2 = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{3n} \frac{1}{\sqrt{n+r}}$$

Factor out $\sqrt{n}$ from the denominator inside the summation to match the Riemann sum form:

$$L_2 = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{3n} \frac{1}{\sqrt{n\left(1+\frac{r}{n}\right)}}$$

$$L_2 = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{3n} \frac{1}{\sqrt{n}\sqrt{1+\frac{r}{n}}}$$

Combine the $\frac{1}{\sqrt{n}}$ terms outside and inside the sum:

$$L_2 = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{3n} \frac{1}{\sqrt{1+\frac{r}{n}}}$$

Identify the integral components:

Function: $f(x) = \frac{1}{\sqrt{1+x}}$

Lower Limit: As $r=1$ and $n \to \infty$, $\frac{r}{n} \to 0$.

Upper Limit: As $r=3n$, $\frac{r}{n} = \frac{3n}{n} \to 3$.

So, the integral is:

$$L_2 = \int_{0}^{3} \frac{1}{\sqrt{1+x}} \, dx = \int_{0}^{3} (1+x)^{-1/2} \, dx$$

Now, integrate:

$$L_2 = \left[ \frac{(1+x)^{1/2}}{1/2} \right]_0^3$$

$$L_2 = 2 \left[ \sqrt{1+x} \right]_0^3$$

$$L_2 = 2 \left( \sqrt{1+3} - \sqrt{1+0} \right)$$

$$L_2 = 2 (\sqrt{4} - \sqrt{1})$$

$$L_2 = 2 (2 - 1) = 2(1) = 2$$

So, $L_2 = 2$

Final Calculation

The total limit $L$ is the sum of the two parts:

$$L = L_1 + L_2$$

$$L = 0.2 + 2$$

$$L = 2.2$$

Rounding off to two decimal places:

Answer = 2.20