Question

$\lim_{n \to \infty} \frac{(1^5+2^5+\dots+n^5)(1^8+2^8+\dots+n^8)}{1^{14}+2^{14}+\dots+n^{14}}$

• A. 5/18

Answer 1

Answer: (A)

5/18

Answer 2

The problem asks us to evaluate the limit of an expression involving sums of powers of natural numbers.

The general formula for the sum of the $p$-th powers of the first $n$ natural numbers is given by: $\sum_{k=1}^{n} k^p = 1^p + 2^p + \dots + n^p$ As $n \to \infty$, the dominant term in this sum is $\frac{n^{p+1}}{p+1}$. More formally, we can state that: $\lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^p}{n^{p+1}} = \frac{1}{p+1}$ This means that for large $n$, $\sum_{k=1}^{n} k^p \approx \frac{n^{p+1}}{p+1}$.

Let's apply this approximation to each sum in the given expression:

1. For the sum $1^5+2^5+\dots+n^5$: Here $p=5$, so the sum is approximately $\frac{n^{5+1}}{5+1} = \frac{n^6}{6}$.

2. For the sum $1^8+2^8+\dots+n^8$: Here $p=8$, so the sum is approximately $\frac{n^{8+1}}{8+1} = \frac{n^9}{9}$.

3. For the sum $1^{14}+2^{14}+\dots+n^{14}$: Here $p=14$, so the sum is approximately $\frac{n^{14+1}}{14+1} = \frac{n^{15}}{15}$.

Now, substitute these approximations into the limit expression: $\lim_{n \to \infty} \frac{(1^5+2^5+\dots+n^5)(1^8+2^8+\dots+n^8)}{1^{14}+2^{14}+\dots+n^{14}} = \lim_{n \to \infty} \frac{\left(\frac{n^6}{6}\right)\left(\frac{n^9}{9}\right)}{\left(\frac{n^{15}}{15}\right)}$

Simplify the expression: $= \lim_{n \to \infty} \frac{\frac{n^{6+9}}{6 \times 9}}{\frac{n^{15}}{15}}$ $= \lim_{n \to \infty} \frac{\frac{n^{15}}{54}}{\frac{n^{15}}{15}}$ To simplify further, we can multiply the numerator by the reciprocal of the denominator: $= \lim_{n \to \infty} \frac{n^{15}}{54} \times \frac{15}{n^{15}}$ The $n^{15}$ terms cancel out: $= \lim_{n \to \infty} \frac{15}{54}$ The limit is a constant value, which is $\frac{15}{54}$.

Finally, simplify the fraction $\frac{15}{54}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 3: $\frac{15 \div 3}{54 \div 3} = \frac{5}{18}$

The final answer is $\frac{5}{18}$.

Explanation of the solution: The problem involves limits of sums of powers. The key is to use the asymptotic behavior of the sum of $p$-th powers of the first $n$ natural numbers, $\sum_{k=1}^{n} k^p \sim \frac{n^{p+1}}{p+1}$ as $n \to \infty$.

1. Replace $1^5+2^5+\dots+n^5$ with its leading term $\frac{n^6}{6}$.
2. Replace $1^8+2^8+\dots+n^8$ with its leading term $\frac{n^9}{9}$.
3. Replace $1^{14}+2^{14}+\dots+n^{14}$ with its leading term $\frac{n^{15}}{15}$.
4. Substitute these into the limit expression: $\lim_{n \to \infty} \frac{\left(\frac{n^6}{6}\right)\left(\frac{n^9}{9}\right)}{\left(\frac{n^{15}}{15}\right)} = \lim_{n \to \infty} \frac{\frac{n^{15}}{54}}{\frac{n^{15}}{15}}$
5. Simplify by canceling $n^{15}$ and evaluating the constant fraction: $\frac{15}{54} = \frac{5}{18}$