Let $\sum_{n=1}^\infty a_n$ be an infinite series with the property that for every positive integer $N$, the $N^{th}$ partial sum of the series is $S_N = \sum_{n=1}^N a_n = 2 - \frac{1}{N}$

a) Does this series converge or diverge? Justify your answer.

I was going to say that $\sum\frac{1}{N}$ is the harmonic series, which diverges, so the whole function diverges, but another part of me thinks that the $\sum a_n$ converges because as $N\to \infty$, $\sum a_n = 2$. Can someone explain which train of logic is correct?

b) Show that $a_n > 0$ for every positive integer $n$.

For $n=1$, $a_n = 1$, and as $n \to \infty, a_n = 2$. Since $1 < a_n < 2$, $a_n$ is positive. Is that the correct way of putting it?

Thank you for any help!


a) You only have $\frac 1 N$ in the partial sums, not in the individual terms of the series. So why would you compare to the harmonic series? Your second line is correct. By definition, a series "converges" if the sequence of partial sums converges. And $S_N$ converges (to 2 as you noted).

b) Where did you get $1 < a_n < 2$ from? If you knew that, you would also know that all terms were strictly positive. Instead, by definition $a_n = S_n - S_{n-1}$; plug in the formula for $S_n$ to compute every $a_n$ explicitly. In particular you will find they are all positive. OR: simply show directly that $S_n > S_{n-1}$ which is the same as $a_n >0$.

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    $\begingroup$ Not OP, but would it be correct to say for part b, that 1/N = 1 when N=1 and 2 - 1 > 0. And since 1/N is monotonically decreasing for N in [1,inf), we can say that 2 - 1/N will always be greater than 0. $\endgroup$ – orbit-stabilizer Jul 13 '16 at 0:41
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    $\begingroup$ Yes. That is the argument missing from the OP's answer to (b). $\endgroup$ – mathguy Jul 13 '16 at 0:45
  • $\begingroup$ a) Thank you kindly! b) So I would say $a_n = (2 - \frac{1}{n}) - (2 - \frac{1}{n-1}) = \frac{1}{n(n-1)}$, which for all positive n is greater than 0? $\endgroup$ – S.Kumar Jul 13 '16 at 0:48
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    $\begingroup$ Correct! ........ (...... added to meet the minimum characters requirement) $\endgroup$ – mathguy Jul 13 '16 at 0:49
  • $\begingroup$ Thank you so much! I'm going to format io_cantor's argument for others to see more clearly (and to also check if I'm writing it correctly: For $\sum_{n=1}^{1} {a_n}, a_n = 2 - \frac{1}{1} = 1$ Since $\frac{1}{N}$ is monotone decreasing for all positive $n$, $2 - \frac{1}{N}$ is bounded below by 1, which is $>0$ $\endgroup$ – S.Kumar Jul 13 '16 at 0:54

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