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Find the values of ##x## for which the following series is convergent.

I compared the series with the harmonic series and deduced it is always divergent. I used ##y^p<e^y## for large ##y##. I used a different method from the answer given, which I don't understand.

When ##k=1##, ##M_1=e=2.72## and ##M_0=1##. Since ##M_0+1\leq n\leq M_1##, we have ##n=2##, giving us the term ##\frac{1}{1^X}##.

When ##k=2##, ##M_2=e^2=7.39## and ##M_1=e=2.72##. Since ##M_1+1\leq n\leq M_2##, we have ##n=4, 5, 6, 7##, giving us the terms ##\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}##.

Finding the possible values of ##n## is troublesome.

Is there a typo in the given answer?

I compared the series with the harmonic series and deduced it is always divergent. I used ##y^p<e^y## for large ##y##. I used a different method from the answer given, which I don't understand.

When ##k=1##, ##M_1=e=2.72## and ##M_0=1##. Since ##M_0+1\leq n\leq M_1##, we have ##n=2##, giving us the term ##\frac{1}{1^X}##.

When ##k=2##, ##M_2=e^2=7.39## and ##M_1=e=2.72##. Since ##M_1+1\leq n\leq M_2##, we have ##n=4, 5, 6, 7##, giving us the terms ##\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}##.

Finding the possible values of ##n## is troublesome.

Is there a typo in the given answer?

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