Math  /  Calculus

QuestionDetermine whether the given series is convergent or divergent. n=1n2n2+1\sum_{n=1}^{\infty} \frac{n}{\sqrt{2 n^{2}+1}} convergent divergent

Studdy Solution

STEP 1

What is this asking? Does this series, which is a sum of infinitely many terms, settle down to a specific value or does it explode to infinity? Watch out! Don't jump to conclusions based on the first few terms!
Infinity is a long time, and things can change dramatically as nn gets **really** big.

STEP 2

1. Analyze the terms
2. Apply the Divergence Test

STEP 3

Let's **look** at the general term of our series: an=n2n2+1a_n = \frac{n}{\sqrt{2n^2 + 1}}.
We want to see what happens when nn becomes **super huge**!

STEP 4

To do this, we'll **divide** both the numerator and the denominator by the highest power of nn in the denominator, which is nn.
Remember, we're doing this because as nn gets **massive**, the highest power terms dominate the expression.

STEP 5

limnan=limnn2n2+1 \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{\sqrt{2n^2 + 1}}

STEP 6

We can rewrite the expression inside the limit as: limnn/n2n2/n2+1/n2 \lim_{n \to \infty} \frac{n/n}{\sqrt{2n^2/n^2 + 1/n^2}}

STEP 7

This simplifies to: limn12+1/n2 \lim_{n \to \infty} \frac{1}{\sqrt{2 + 1/n^2}}

STEP 8

As nn approaches infinity, 1/n21/n^2 approaches **zero**.
So, our limit becomes: limn12+0=12 \lim_{n \to \infty} \frac{1}{\sqrt{2 + 0}} = \frac{1}{\sqrt{2}}

STEP 9

The Divergence Test tells us that if the limit of the terms of a series as nn goes to infinity is *not* zero, then the series **diverges**.

STEP 10

We just found that the limit of our terms is 12\frac{1}{\sqrt{2}}, which is definitely *not* zero!

STEP 11

Therefore, by the Divergence Test, our series n=1n2n2+1\sum_{n=1}^{\infty} \frac{n}{\sqrt{2 n^{2}+1}} **diverges**!

STEP 12

The series is **divergent**.

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