For every positive integer $n$, let $p(n)$ denote the number of ways to express $n$ as a sum of positive integers. For instance, $p(4)=5$. Also define $p(0)=1.$
Problem 1. Prove that $p(n)-p(n-1)$ is the number of ways to express $n$ as a sum of integers each of which is strictly greater than $1$.
Problem 2. Prove that the total number of components in all partitions of $n$ is equal to $\displaystyle\sum_{k=0}^{n-1}p(k)\cdot \tau(n-k)$, where $\tau(m)=\sum_{d|m} 1$ is the number of positive divisors of $m$.
I came up with a solution for first problem, but I'm not sure if everything is OK with it, so I would be very grateful if you tell me what I should correct here:
Let $p_k(n)$ denote number of ways to express $n$ as a sum of exactly $k$ positive integers. So $p_k(n)=\left|\left\{ \langle a_1,...,a_k \rangle : \sum_{i=1}^k a_i=n \wedge 1\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\}\right|$. It's quite easy to observe that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ by dividing above set into two disjoint sets:
-set of tuples $\langle 1,...,a_k \rangle$
-set of tuples $\langle a_1,...,a_k \rangle$, where $a_1>1$
So let's count $p(n)-p(n-1)$ knowing that $p(n)=\sum_{k=1}^n p_k(n)$.
$p(n)-p(n-1)=p_n(n)+\sum_{k=1}^{n-1}p_k(n)-p_k(n-1)=\\=1+\sum_{k=1}^{n-1}p_{k-1}(n-1)+p_k(n-k)-p_k(n-1)=\\=1+ p_0(n-1)+p_1(n-1)-p_1(n-1)+p_1(n-1)+p_2(n-2)-p_2(n-1)+\dots+p_{n-3}(n-1)+p_{n-2}(2)-p_{n-2}(n-1)+p_{n-2}(n-1)+p_{n-1}(1)-p_{n-1}(n-1)=\\=1+0-p_{n-1}(n-1)+\sum_{k=1}^{n-1}p_k(n-k)=\sum_{k=1}^{n-1}p_k(n-k)$
On the other hand, let $r(n)$ be number of partitions that we are looking for: $r(n)=\left| \left\{ \langle a_1,...,a_k \rangle : 1\le k\le n-1, \ \sum_{i=1}^k a_i=n \wedge 2\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\} \right|$
and for a given $k$:
$r_k(n)=\left| \left\{ \langle a_1,...,a_k \rangle : \sum_{i=1}^k a_i=n \wedge 2\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\} \right|$ is a number of partitions that we are looking for but consisting only of $k$ integers.
Since there is a simple bijection between tuples $\langle a_1,...,a_k \rangle \leftrightarrow \langle a_1-1,...,a_k-1 \rangle $, where $2\le a_i\le a_{i+1}$ for $1\le i\le k-1$ and still $1\le a_i-1$, then $\sum_{i=1}^k(a_i-1)=n-k$ and tuple $\langle a_1-1,...,a_k-1 \rangle$ is some partition of $n-k$. So $r_k(n)=p_k(n-k)$.
Hence cases $r_k(n)$ are pairwise disjoint for $1\le k\le n-1$, it follows that: $\displaystyle r(n)=\sum_{k=1}^{n-1}r_k(n)=\sum_{k=1}^{n-1}p_k(n-k)=p(n)-p(n-1)$ as desired.
Is that correct?
I don't know how to approach second problem. Too hard it seems to be.