Cfrgtkky

An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):

2 points in [1, 2):

And so on:

Now you have a sequence that grows to infinity but keeps getting closer together.

Any increasing sequence ${a_n}_{ngeq 1}$ has limit in $mathbb{R}cup{+infty}$. It is $sup_{ngeq 1} a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_{n+1}-a_nto 0$).

An example with a finite limit is $a_n=1-1/nto 1$ and $a_{n+1}-a_n=frac{1}{n(n+1)}to 0$.

On the other hand $a_n=sqrt{n}to +infty$ and $a_{n+1}-a_n=sqrt{n+1}-sqrt{n}=frac{1}{sqrt{n+1}+sqrt{n}}to 0$.

So, the answer is NO, the condition $a_{n+1}-a_nto 0$ is not sufficient for an increasing sequence ${a_n}_{ngeq 1}$ to have a FINITE limit.

Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.

No. Consider the sequence ${a_n}_{n=1}^infty$ given by

• 
  $a_n = sumlimits_{k=1}^{n} frac{1}{k}$.

It follows that

• $a_n > a_{n-1}$


• 
  $a_n - a_{n-1} = frac{1}{n} rightarrow 0$ as $n rightarrow infty$, but


• 
  $a_n = sumlimits_{k=1}^{n} frac{1}{k} rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).

The condition $a_{n+1}-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_{ntoinfty}(a_{n+m(n)}-a_n)=0$ for all $m(n)in mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.

Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.

Note that if we define $b_n=a_{n+1}-a_n$, then $a_n=a_0+sum_{n=0}^{infty}b_n$. So this question is equivalent to asking whether the terms of an infinite series going to zero is sufficient for the series to converge. There are a variety of examples of series with terms that go to zero, yet do not converge, with the harmonic series ($sum frac 1 n$) being one of the most famous.

And in fact we can construct a counterexample from any sequence by defining a sequence $c_n$ by simply re-indexing the terms. We set $c_0$ equal to $a_0$. Then set $c_{k1}$ equal to $a_1$, where $k_1>a_1-a_0$, and fill in the terms $c_1$ to $c_{k-1}$ with equally spaced terms; this will result in all of the consecutive differences from $c_0$ to $c_{k1}$ being less than $1$. Then set $c_k2$ equal to $a_2$, where $k_2+k_1>2(a_2-a_1)$, which results in consecutive differences between $c_{k1}$ to $c_{k2}$ being less than $frac 1 2$. Just keep re-indexing each term and filling in more and more new terms, and you can drive the consecutive differences arbitrarily low.

Another approach is to look at a sequence as an approximation of a continuous function, and the difference between successive terms as an approximation of the derivative. Then we just need a function such that $f'(x)$ converges to zero, but $f$ diverges. Two examples of such are the log function, (which gives a sequence very similar to the harmonic sequence) and square root. Note that both of these examples can be obtained by taking the inverse of a function whose derivative is constantly increasing. If $g'$ goes to infinity, then $(g^{-1})'$ goes to zero. But if the domain of $g$ is the whole real line, then the range of $g^{-1}$ is the whole real line, i.e. $g^{-1}$ goes to infinity.