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------------------------------------------------ Use theorem 9. 11 to determine the convergence or divergence of the p-series. 1 + 1 32 2 + 1 243 3 + 1 1024 4 + 1 3125 5 +. . .

Answer :

According to Theorem 9.11, since p ≤ 1, the p-series Σ(1/nᵖ) diverges. This means that the given series 1 + 1/32 + 1/243 + 1/1024 + 1/3125 + ... also diverges.

The given series is a p-series of the form Σ(1/nᵖ), where n is the term number and p is a constant exponent.

To determine the convergence or divergence of the series, we can use Theorem 9.11, which states:

If p > 1, then the p-series Σ(1/nᵖ) converges.
If p ≤ 1, then the p-series Σ(1/nᵖ) diverges.

In the given series, the terms are 1, 1/32, 1/243, 1/1024, 1/3125, and so on.

Notice that the terms in the series are of the form 1/nᵖ, where n represents the term number and p is a constant exponent. In this case, the exponent p is equal to 1, so we have a p-series with p ≤ 1.

According to Theorem 9.11, since p ≤ 1, the p-series Σ(1/nᵖ) diverges.

This means that the given series 1 + 1/32 + 1/243 + 1/1024 + 1/3125 + ... also diverges.

Therefore, the given series diverges.

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