Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is writing a sequence as a telescoping sum the only way to turn a sequence into an infinite series? In particular:

Let $s_n = 1 +\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}- \ln(n+1), \ n \geq 1$. Convert this into an infinite series.

So let $a_n = s_n-s_{n-1}$. Then $a_n = \frac{1}{n}-\ln(n+1)+ \ln n$. This is equivalent to $$\sum_{n=1}^{\infty} \left(\frac{1}{n}- \ln \frac{n+1}{n} \right)$$

What does the inside term represent?

share|cite|improve this question
What do you mean what does it represent? There's no new notation or terminology there, and I assume you know what a logarithm is. Maybe have a look at – anon Jul 4 '11 at 21:17
You don't just need to take a limit, hmm? – mixedmath Jul 4 '11 at 21:33
I am afraid I do not even know what the question is. – Did Jul 4 '11 at 21:46

The term $\dfrac{1}{n}- \ln \dfrac{n+1}{n}$ represents this mauve area

enter image description here

The limit of the sum is the Euler–Mascheroni constant, about $0.577\ldots$

share|cite|improve this answer

Is the "inside term" $$\frac{1}{n}- \ln \frac{n+1}{n}?$$

If that is the case, there is a clear geometric interpretation. Draw the curve $y =1/x$. Draw also the following rectangles. They all have width $1$. The leftmost one has lower left corner at $(1,0)$, height $1$. The next one has lower left corner at $(2,0)$, height $1/2$. The next has lower left corner at $(3,0)$, height $1/3$. And so on.

When a partial sum of the harmonic series is approximated by a suitable definite integral of $1/x$, the "inside terms" represent the local approximation error, the error obtained by approximating the area of the rectangle by the area under $y=1/x$.

Added: In a not very interesting sense, every (convergent) series can be viewed as a telescoping sum. Consider the series $\sum_{n=1}^\infty a_n$. Let $s_n=\sum_{k=1}^n a_k$, and for convenience put $s_0=0$. Then $$\sum_{n=1}^\infty a_n=(s_1-s_0)+(s_2-s_1)+(s_3-s_2) +\cdots.$$

In particular, any sequence can be analyzed as coming from a telescoping series. If the sequence is to be reproduced term by term, then the obvious telescoping series representation is unavoidable. But there are many sophisticated ways of accelerating convergence that do not reproduce the sequence term by term, and for which a "telescoping series" description is not useful.

share|cite|improve this answer

A sequence and a series is pretty much the same object with different points of view. When we talk about "the series $\sum_{n \ge 1} u_n$", it usually means we're interested in "the sequence $\left(\sum_{k = 0}^n u_k \right)_{n \ge 1}$", but we talk about a series instead because defining a sequence using the series notation is sometimes convenient. We can talk about a series regardless of convergence (just like there are non convergent sequences). In fact, there is a bijection between sequences and series (again, I'm not assuming any kind of convergence here). In one way, you go form a sequence to a series (the "telescoping sum") by :

$$(s_n)_{n \ge 1} \longmapsto \sum_{n \ge 1} s_n - s_{n_1}$$

On the other way, the inverse process from a series to a sequence ("partial sums") is

$$\sum_{n \ge 1} a_n \longmapsto \left(\sum_{k = 0}^n a_k \right)_{n \ge 1}$$

Moreover, this bijection induces one between convergent series and convergent sequences (and the limit of the sequence of partial sums is what is called the sum of the series and is denoted $\sum_{n =1}^{\infty} a_n$).

In your example, it can be useful to write the inside term as

$$a_n = \frac{1}{n} - \ln \frac{n+1}{n} = \int_{n}^{n+1} \left(\frac{1}{n} - \frac{1}{t}\right) \ dt$$

And what we denote as $\sum_{n =1}^{\infty} a_n$ is $\lim_{n \to \infty} s_n$.

share|cite|improve this answer

It may be worth noting the following. $$ s_n = \sum\limits_{k = 1}^n {\frac{1}{k}} - \int_1^{n + 1} {\frac{1}{x}\,dx} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \sum\limits_{k = 1}^n {\int_k^{k + 1} {\frac{1}{x}\,dx} } = \sum\limits_{k = 1}^n {\bigg(\frac{1}{k} - \int_k^{k + 1} {\frac{1}{x}\,dx} \bigg)} . $$ Hence $s_n = \sum\nolimits_{k = 1}^n {a_k } $ with $a_n \,(= s_n - s_{n-1})$ given by $$ a_n = \frac{1}{n} - \int_n^{n + 1} {\frac{1}{x}\,dx} = \frac{1}{n} - \ln \frac{{n + 1}}{n}. $$ Now, from the simple fact that, for any $x > 0$, $$ 0 < x - \ln (1 + x) < x^2 $$ it follows that, for any $n \in \mathbb{N}$, $$ 0 < \frac{1}{n} - \ln \bigg(1 + \frac{1}{n}\bigg) < \frac{1}{{n^2 }}, $$ so $$ 0 < a_n < \frac{1}{{n^2 }}. $$ Hence $$ \sum\limits_{n = 1}^\infty {\bigg(\frac{1}{n} - \ln \frac{{n + 1}}{n}\bigg)} = \sum\limits_{n = 1}^\infty {a_n } < \infty . $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.