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Determine the value of [tex]\( p \)[/tex] for the [tex]\( p \)[/tex]-series given below.

[tex]\[
\sum_{n=1}^{\infty} \frac{3}{n^2 \cdot n^{1 / 3}}
\][/tex]

Sagot :

GinnyAnsweridn
To determine the value of [tex]\( p \)[/tex] for the given [tex]\( p \)[/tex]-series, we start by analyzing the series:

[tex]\[ \sum_{n=1}^{\infty} \frac{3}{n^2 \cdot n^{1/3}} \][/tex]

First, we simplify the denominator. The expression [tex]\( n^2 \cdot n^{1/3} \)[/tex] can be rewritten using the properties of exponents:

[tex]\[ n^2 \cdot n^{1/3} = n^{2 + 1/3} \][/tex]

This simplification gives us:

[tex]\[ n^{2 + 1/3} = n^{\frac{6}{3} + \frac{1}{3}} = n^{\frac{7}{3}} \][/tex]

Now, we substitute this back into the series:

[tex]\[ \sum_{n=1}^{\infty} \frac{3}{n^{7/3}} \][/tex]

Next, we recall the general form of a [tex]\( p \)[/tex]-series:

[tex]\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \][/tex]

By comparing the simplified series with the general form, we observe that [tex]\( p \)[/tex] in our case is the exponent in the denominator's [tex]\( n \)[/tex]-term. For the given series:

[tex]\[ \sum_{n=1}^{\infty} \frac{3}{n^{7/3}} \][/tex]

we can see that [tex]\( p = \frac{7}{3} \)[/tex].

Converting the fraction [tex]\(\frac{7}{3} \)[/tex] to a decimal, we get:

[tex]\[ \frac{7}{3} \approx 2.3333333333333335 \][/tex]

Hence, the value of [tex]\( p \)[/tex] for the given [tex]\( p \)[/tex]-series is:

[tex]\[ \boxed{2.3333333333333335} \][/tex]
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