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: Recall the sum formulas.
$1 + 2 + \ldots + n = \frac{n(n+1)}{2}$ and $1^2 + 2^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.

Step 2: Set up $a$.
\[ a = \lim_{n\to\infty} \frac{n(n+1)}{2n^2} = \lim_{n\to\infty} \frac{n^2+n}{2n^2} \]
Step 3: Take the limit for $a$.
Divide top and bottom by $n^2$. The $\frac{1}{n}$ term goes to 0.
\[ a = \frac{1}{2} \]
Step 4: Set up $b$.
\[ b = \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{6n^3} = \lim_{n\to\infty} \frac{2n^3 + 3n^2 + n}{6n^3} \]
Step 5: Take the limit for $b$.
Divide by $n^3$; only the leading term survives.
\[ b = \frac{2}{6} = \frac{1}{3} \]
Step 6: Compare $a$ and $b$.
$a = \frac{1}{2}$ gives $2a = 1$. $b = \frac{1}{3}$ gives $3b = 1$. So $2a = 3b$. \[ \boxed{2a = 3b \text{ (Option 2)}} \]