Question

If $a = \lim_{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ and $b = \lim_{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$, then

• A. $a = b$
• B. $2a = 3b$
• C. $a = 2b$
• D. $3a = 2b$

Hint:

For limits approaching infinity of polynomials as a rational function, simply look at the ratio of the coefficients of the highest power of $n$. For $a$, it's $1/2$. For $b$, the highest term in the numerator expansion is $2n^3$, so the ratio is $2/6 = 1/3$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate two separate infinite limits involving the sum of the first $n$ natural numbers and the sum of their squares, and then find the relation between the results $a$ and $b$.

Step 2: Key Formula or Approach:
The sum of the first $n$ natural numbers is $\sum n = \frac{n(n+1)}{2}$.
The sum of the squares of the first $n$ natural numbers is $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.

Step 3: Detailed Explanation:
First, calculate the limit for $a$:
$$a = \lim_{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{n^2} = \lim_{n \rightarrow \infty} \frac{n^2 + n}{2n^2}$$
Divide the numerator and denominator by $n^2$:
$$a = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1 + 0}{2} = \frac{1}{2}$$
Next, calculate the limit for $b$:
$$b = \lim_{n \rightarrow \infty} \frac{\frac{n(n+1)(2n+1)}{6}}{n^3} = \lim_{n \rightarrow \infty} \frac{2n^3 + 3n^2 + n}{6n^3}$$
Divide the numerator and denominator by $n^3$:
$$b = \lim_{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2 + 0 + 0}{6} = \frac{2}{6} = \frac{1}{3}$$
Now we have $a = \frac{1}{2}$ and $b = \frac{1}{3}$.
Cross-multiplying these gives $2a = 1$ and $3b = 1$.
Therefore, $2a = 3b$.

Step 4: Final Answer:
The relationship is $2a = 3b$, matching option (B).