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To test the series $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ for convergence, you can use the P-test. (You could also use the Integral Test, as is the case with all series of this type.) According to the P-test: $\circ \sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ diverges $\circ \sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ converges $\circ$ the P-test does not apply to $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ Now compute $s_4$, the partial sum consisting of the first 4 terms of $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$: $s_4 = $

          To test the series $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ for convergence, you can use the P-test. (You could also use the Integral Test, as is the case with all series of this type.) According to the P-test:
$\circ \sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ diverges
$\circ \sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$ converges
$\circ$ the P-test does not apply to $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$
Now compute $s_4$, the partial sum consisting of the first 4 terms of $\sum_{k=1}^{\infty} \frac{1}{\sqrt[5]{k^5}}$:
$s_4 = $

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To test the series ∑k=1^∞(1)/(√(k^5)) for convergence, you can use the P-test. (You could also use the Integral Test, as is the case with all series of this type.) According to the P-test:
∘∑k=1^∞(1)/(√(k^5)) diverges
∘∑k=1^∞(1)/(√(k^5)) converges
∘ the P-test does not apply to ∑k=1^∞(1)/(√(k^5))
Now compute s4, the partial sum consisting of the first 4 terms of ∑k=1^∞(1)/(√(k^5)):
s4 =

Added by Jeffrey I.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Manisha Sarker

08/23/2023

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In this case, the series is not given, so we cannot apply the P-test. Show more…

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To test the series for convergence, you can use the P-test. (You could also use the Integral Test, as is the case with all series of this type.) According to the P-test: 1 diverges, VRS 1 converges, V7.5 The P-test does not apply to this series. Now compute √54, the partial sum consisting of the first 4 terms.

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