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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