Calculate the limit of a sequence $a_n = \frac{(1+3+5+...+(2n-1)}{(n^2+1)}$

Question

Calculate the limit of a sequence $a_n$, where $$a_n = \frac{1+3+5+\cdots+(2n-1)}{n^2+1}$$

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MY TRY: $$\lim_{n\to\infty}\frac{2n-1}{n^2+1} $$

how would you solve this, did I even set the limit up correctly?

Answer 1

Hint:

Computing the first few terms of the numerator is helpful:

$$1,4,9,16,25,36,\cdots$$

Answer 2

Hint:

Sum of first $n$ odd natural numbers is $n^2$

$$\lim_{n\to\infty} a_n =\lim_{n\to\infty} \frac {n^2}{n^2+1} =1$$

Answer 3

Limit value = $1$.

Add and subtract $2,4,6,..., 2n$ in the numerator.

After applying sum of first n natural numbers formula, then numerator becomes $n^2$.

Formula: $1+2+3+...+n=n(n+1)/2$

Answer 4

$$a_n = \frac{1+3+5+7+...+(2n-1)}{n^2+1}$$ $$\lim_{n \to \infty} a_n = ?$$ Well, first, look at the numerator. $$1=1$$ $$1+3 = 4$$ $$1+3+5 = 9$$ $$1+3+5+7 = 16$$ $$1+3+5+7+9 = 25$$ As you can see, adding up consecutive odd numbers gives perfect squares. But, you can notice a general pattern if you pay more attention to the numbers. $$1 = \biggr(\frac{1+1}{2}\biggr)^2$$ $$4 = \biggr(\frac{3+1}{2}\biggr)^2$$ $$9 = \biggr(\frac{5+1}{2}\biggr)^2$$ $$16 = \biggr(\frac{7+1}{2}\biggr)^2$$ $$25 = \biggr(\frac{9+1}{2}\biggr)^2$$

Applying this pattern to the first, we can simplify the numerator of the limit.

$$1+3+5+7+...+(2n-1) = \biggr(\frac{(2n-1)+1}{2}\biggr)^2$$ $$1+3+5+7+...+(2n-1) = \biggr(\frac{2n}{2}\biggr)^2 = n^2$$

So, rewrite the limit now.

$$\lim_{n \to \infty} \frac{n^2}{n^2+1}$$

When you have a rational function in the form $y = \frac{P(x)}{Q(x)}$, if $P(x)$ and $Q(x)$ have the same degree, divide the coefficients of the leading terms to get the limit as $x \to \infty$. Both coefficients are $1$. Therefore, the limit becomes $1$.

$$\boxed{\lim_{n \to \infty} \frac{n^2}{n^2+1} = 1}$$