Studdy AI Logo
Math  /  Calculus

Question∑n=1∞n3n5=1n2\sum_{n=1}^{\infty} \frac{n^{3}}{n^{5}}=\frac{1}{n^{2}}
Converge ∑n=1∞n3−2n2+n−1n5−2\sum_{n=1}^{\infty} \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2}

Studdy Solution

STEP 1

What is this asking? Does the infinite sum ∑n=1∞n3−2n2+n−1n5−2\sum_{n=1}^{\infty} \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2} converge? Watch out! Don't forget to check the conditions for convergence tests carefully!

STEP 2

1. Simplify the Summand
2. Limit Comparison Test

STEP 3

Let's **look** at our sum: ∑n=1∞n3−2n2+n−1n5−2\sum_{n=1}^{\infty} \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2}.
It looks a little complicated, right?
So let's think about what happens when nn gets **really big**!

STEP 4

When nn is **huge**, the highest power terms dominate.
So, n3n^3 is *way* bigger than 2n22n^2, nn, and 11.
Similarly, n5n^5 is *much* larger than 22.

STEP 5

That means our summand behaves *like* n3n5=1n2\frac{n^3}{n^5} = \frac{1}{n^2} for **large** nn.
This gives us a great idea for what to compare it to!

STEP 6

We'll use the **Limit Comparison Test**.
Let an=n3−2n2+n−1n5−2a_n = \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2} and bn=1n2b_n = \frac{1}{n^2}.
We want to find lim⁡n→∞anbn\lim_{n\to\infty} \frac{a_n}{b_n}.

STEP 7

Let's **calculate** that limit: lim⁡n→∞anbn=lim⁡n→∞n3−2n2+n−1n5−21n2\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{n^{3}-2 n^{2}+n-1}{n^{5}-2}}{\frac{1}{n^2}}

STEP 8

To **simplify**, we can multiply by n2n2\frac{n^2}{n^2} (which is equivalent to multiplying by one): lim⁡n→∞n2(n3−2n2+n−1)n5−2=lim⁡n→∞n5−2n4+n3−n2n5−2\lim_{n\to\infty} \frac{n^2(n^{3}-2 n^{2}+n-1)}{n^{5}-2} = \lim_{n\to\infty} \frac{n^{5}-2 n^{4}+n^{3}-n^2}{n^{5}-2}

STEP 9

Now, we can **divide** both the numerator and denominator by n5n^5 (which is equivalent to dividing by one): lim⁡n→∞1−2n+1n2−1n31−2n5\lim_{n\to\infty} \frac{1-\frac{2}{n}+\frac{1}{n^{2}}-\frac{1}{n^{3}}}{1-\frac{2}{n^{5}}}

STEP 10

As nn goes to infinity, all terms with nn in the denominator go to **zero**.
So, the limit is 1−0+0−01−0=11=1\frac{1-0+0-0}{1-0} = \frac{1}{1} = \mathbf{1}.

STEP 11

Since the limit is 1\mathbf{1} (a positive, finite number), and we know that ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2} converges (it's a p-series with p=2>1p=2 > 1), then by the **Limit Comparison Test**, our original series ∑n=1∞n3−2n2+n−1n5−2\sum_{n=1}^{\infty} \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2} also **converges**!

STEP 12

The series ∑n=1∞n3−2n2+n−1n5−2\sum_{n=1}^{\infty} \frac{n^{3}-2 n^{2}+n-1}{n^{5}-2} converges.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord