Studdy AI Logo
Math  /  Calculus

Question3. ∑n=1∞5n3n4+7\sum_{n=1}^{\infty} \frac{5 n^{3}}{n^{4}+7}

Studdy Solution

STEP 1

1. We are dealing with an infinite series ∑n=1∞5n3n4+7\sum_{n=1}^{\infty} \frac{5n^3}{n^4+7}.
2. The goal is to determine the convergence or divergence of the series.
3. We will use the Limit Comparison Test to analyze the series.

STEP 2

1. Identify a suitable comparison series.
2. Apply the Limit Comparison Test.
3. Determine the convergence or divergence of the original series based on the test.

STEP 3

Identify a suitable comparison series. We observe that for large n n , the term 5n3n4+7\frac{5n^3}{n^4+7} behaves like 5n3n4=5n\frac{5n^3}{n^4} = \frac{5}{n}.
Thus, we compare it with the harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}, which is known to diverge.

STEP 4

Apply the Limit Comparison Test. We calculate the limit:
lim⁡n→∞5n3n4+71n=lim⁡n→∞5n4n4+7\lim_{n \to \infty} \frac{\frac{5n^3}{n^4+7}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{5n^4}{n^4+7}
Simplify the expression:
=lim⁡n→∞5n4n4(1+7n4)=lim⁡n→∞51+7n4=5= \lim_{n \to \infty} \frac{5n^4}{n^4(1 + \frac{7}{n^4})} = \lim_{n \to \infty} \frac{5}{1 + \frac{7}{n^4}} = 5
Since the limit is a positive finite number (5), the Limit Comparison Test tells us that the original series ∑n=1∞5n3n4+7\sum_{n=1}^{\infty} \frac{5n^3}{n^4+7} behaves like the harmonic series.

STEP 5

Determine the convergence or divergence of the original series. Since the harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n} diverges and the Limit Comparison Test shows that our series behaves like the harmonic series, the original series ∑n=1∞5n3n4+7\sum_{n=1}^{\infty} \frac{5n^3}{n^4+7} also diverges.
The series ∑n=1∞5n3n4+7\sum_{n=1}^{\infty} \frac{5n^3}{n^4+7} diverges.

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