How to determine if a series oscillates ∑(n2+1n2+2)nxn where x is real number.

Step 1 It does the same thing as when x≥1, except it oscillates while it diverges. You can think of it as ∑(n2+1n2+2)n(−1)n|x|n

Step 1 Here, we shall study the absolute convergence because x∈R. Si, let: an=|(n2+1n2+2)nxn| We can use, now, the root-cryteria: limn→+∞ann=limn→+∞|n2+1n2+2×x|∼limn→+∞|x| The root criteria ensure that ∑an converges if and only if |x|&lt;1⇒−1&lt;x&lt;1 Now, we can consider x≥1∨x≤−1. The necessary condition for the convergence is not satisfied, so the series diverges. In conclusion: 1) ∑n=0+∞(n2+1n2+2)nxn converges if and only x∈(−1,1) 2) ∑n=0+∞(n2+1n2+2)nxn diverges if and only if x≤−1∨x≥1

Step 1 It does the same thing as when x≥1, except it oscillates while it diverges. You can think of it as ∑(n2+1n2+2)n(−1)n|x|n

Step 1 Here, we shall study the absolute convergence because x∈R. Si, let: an=|(n2+1n2+2)nxn| We can use, now, the root-cryteria: limn→+∞ann=limn→+∞|n2+1n2+2×x|∼limn→+∞|x| The root criteria ensure that ∑an converges if and only if |x|&lt;1⇒−1&lt;x&lt;1 Now, we can consider x≥1∨x≤−1. The necessary condition for the convergence is not satisfied, so the series diverges. In conclusion: 1) ∑n=0+∞(n2+1n2+2)nxn converges if and only x∈(−1,1) 2) ∑n=0+∞(n2+1n2+2)nxn diverges if and only if x≤−1∨x≥1

Key to "How to determine if a series oscillates ∑(n2+1n2+2)nxn..."

Marenonigt

Answered question

2022-01-15

How to determine if a series oscillates

\sum {(\frac{n^{2} + 1}{n^{2} + 2})}^{n} x^{n}

where x is real number.

Answer & Explanation

Lynne Trussell

Beginner2022-01-16Added 32 answers

Step 1
It does the same thing as when

x \geq 1

, except it oscillates while it diverges. You can think of it as

\sum {(\frac{n^{2} + 1}{n^{2} + 2})}^{n} {(- 1)}^{n} {| x |}^{n}

zurilomk4

Beginner2022-01-17Added 35 answers

Step 1
Here, we shall study the absolute convergence because $x \in R$ . Si, let:
$a_{n} = | (\frac{n^{2} + 1}{n^{2} + 2})^{n} x^{n} |$
We can use, now, the root-cryteria:
$lim_{n \to + \infty} \sqrt[n]{a_{n}} = lim_{n \to + \infty} | \frac{n^{2} + 1}{n^{2} + 2} \times x | \sim lim_{n \to + \infty} | x |$
The root criteria ensure that $\sum a_{n}$ converges if and only if $| x | < 1 \Rightarrow - 1 < x < 1$
Now, we can consider $x \geq 1 \lor x \leq - 1$ . The necessary condition for the convergence is not satisfied, so the series diverges.
In conclusion:
1) $\sum_{n = 0}^{+ \infty} {(\frac{n^{2} + 1}{n^{2} + 2})}^{n} x^{n}$ converges if and only $x \in (- 1, 1)$
2) $\sum_{n = 0}^{+ \infty} {(\frac{n^{2} + 1}{n^{2} + 2})}^{n} x^{n}$ diverges if and only if $x \leq - 1 \lor x \geq 1$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Analysis

Didn't find what you were looking for?