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$$ \sum_{n=1}^\infty \frac{\sqrt{n}}{\sqrt{n^3}-i} $$

I determined that series diverges, because it's less than $\frac{1}{n}$ (I assumed that $i$ has no influence here) and $\frac{1}{n}$ diverges, so, by Comparison test, given series diverges. But I am not sure about this. Am I correctly assuming that imaginary number has no influence in this case?

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2 Answers 2

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$$\sum_{n=1}^\infty \frac{\sqrt{n}}{\sqrt{n^3}-i}$$ write the term as: $$\frac{\sqrt{n}(\sqrt{n^3}+i)}{\sqrt{n^3}-i(\sqrt{n^3}+i)}=\frac{n^2+\sqrt{n}i}{{n^3}+1}$$
now as n is large you get:$$\frac{n^2+\sqrt{n}i}{{n^3}+1}\approx\frac{n^2}{n^3}=\frac{1}{n}$$ which diverges

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Firstly, this is a series of complex numbers so if we must be absolutely precise, then we should say that the modulus of the $n$th term in the series is less than $\frac{1}{n}$.

Secondly, and more importantly, it is not possible to conclude that a series is divergent if it is (term-by-term) less than a divergent series (even if we are talking about series with positive terms). To use the comparison test for divergence, you need to know that the series is (term-by-term) more than a divergent series. (For example, $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges although it is (term-by-term) less than the divergent series $\sum_{n=1}^{\infty} \frac{1}{n}$.)

However, I should add that the series does diverge. Indeed, the $n$th term of the series is $\approx$ $\frac{1}{n}$ in modulus if $n$ is sufficiently large (this can me made precise) and the harmonic series diverges.

Hope this helps!

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